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A 


TREATISE 

01? 

SHADES  AND  SHADOWS, 

AND 

LINEAR  PERSPECTIVE. 


BY  CHARLES  DAVIES, 

PROFESSOR  OF  MATHEMATICS  IN  THE  MILITARY  ACADEMY, 

AND 

AUTHOR  OF  THE  COMMON  SCHOOL  ARITHMETIC,  DESCRIPTIVE  GEOMETRY, 
AND  ELEMENTS  OF  SURVEYING. 


SECOND  EDITIOJS[. 


PUBLISHED  BY 

A.  S.  BARNES  & CO,,  Hartford.— WILEY  & PUTNAM:  COLLINS, 
KEESE  & CO.,  New-York.— PERKINS  & MARVIN,  Boston.— 
THOMAS,  COWPERTHWAIT  & CO.,  Philadelphia.— 
CUSHING  & SONS,  Baltimore. 


1839. 


PREFACE. 


Within  the  last  few  years  a great  change  has  taken 
place  in  public  sentiment,  as  to  the  importance  which 
should  be  given  to  mathematical  studies  in  comparison 
with  other  branches  of  education.  Until  recently  it  was 
thought  that  mere  practical  rules,  unaccompanied  by 
demonstration,  were  abundantly  sufficient  for  all  useful 
applications  of  mathematical  science,  and  that  the  mind 
of  the  scholar  could  find  richer  nutriment  in  Virgil  and 
Homer,  than  in  the  propositions  of  Euclid,  or  the  sublime 
theories  of  Newton. 

But  it  is  auspicious  to  the  cause  of  sound  learning, 
that  these  opinions  have  given  place  to  more  rational 
views  of  education  ; that  we  are  at  length  convinced  it 
is  better  to  reason  than  merely  to  remember ; and  that 
the  value  of  an  education  is  to  be  estimated  by  the 
ability  which  it  gives  to  the  mind  of  thinking  profoundly 
and  reasoning  correctly. 

In  presenting  to  the  public  the  following  Treatise  on 
Shadows  and  Perspective,  the  author  cannot  but  flatter 
himself  that  he  shall  add  something  to  the  common 
stock  of  useful  knowledge.  The  subjects  treated  of 
are  certainly  useful : to  the  architect  and  draftsman  a 

A2 


IV 


PREFACE, 


knowledge  of  them  is  indispensable.  To  find  with  ma 
thematical  accuracy  the  lines  of  shade  and  shadow  on  a 
complicated  building, — which  parts  are  to  be  darkened, 
and  which  parts  are  to  be  made  light  in  a drawing  of  it, 
is  certainly  a difficult  problem  unless  it  be  solved  on 
scientific  principles. 

The  art  of  Perspective  teaches  us  how  to  represent 
on  a surface  one  or  more  solid  bodies,  in  such  a manner 
that  the  picture  shall  exhibit  the  same  appearance  as  is 
presented  by  the  objects  themselves.  It  is  by  this  art 
that  the  painter  is  enabled  to  present  to  the  eye  the 
almost  living  landscape,  with  its  hills,  its  valleys,  its 
waterfalls,  and  its  rich  foliage,  varied  by  the  beautiful 
tints  of  colour,  and  relieved  by  alternate  light  and  shade. 
It  is  this  art  which  has  stamped  the  canvass  with  the 
intelligence  of  the  human  countenance,  and  caused  it  to 
be  looked  upon  as  the  remembrancer  of  departed  worth 
and  the  record  of  former  times.  It  is  this  art  which 
presents  in  a panorama  a city  in  all  its  proportions, 
and  causes  the  spectator  to  feel  that  he  almost  partici- 
pates in  its  bustle  and  business.  This  art  also  enlarges 
the  pleasures  of  sight,  the  sense  through  which  the 
mind  receives  the  most  numerous  and  pleasing  impres- 
sions. 

Without  perfect  accuracy  in  the  perspective,  the  pro- 
portions of  objects  cannot  be  preserved ; and  the  skill 
of  the  draftsman,  or  the  genius  of  the  painter,  is  exerted 
in  vain,  if  nature  be  not  correctly  copied. 


PREFACE. 


V 


The  manner  of  finding  the  shadows  of  objects,  and 
the  common  methods  of  perspective,  depend  on  mathe- 
matical principles,  and  are  susceptible  of  demonstration. 
The  want  of  a work  demonstrating  those  principles 
rigorously,  has  long  been  felt,  and  especially  in  the 
Military  Academy,  where  the  subjects  have  been  taught 
by  lecture  for  several  years.  If  the  one  now  submitted 
be  found  not  to  merit  the  approbation  of  the  public,  the 
author  hopes  it  will  at  least  be  received  with  indulgence. 

The  author  would  beg  leave  to  express  his  acknow- 
ledgments to  two  or  three  friends,  to  whom  he  is  indebted 
for  drafts  of  the  diagrams ; and  also,  to  repeat  his  ex- 
pressions of  thankfulness  to  the  Cadets  for  the  interest 
they  have  taken  in  the  work.  But  for  their  liberality  it 
could  not  have  appeared 

Military  Academy,  ) 

West  Point,  March,  1832.  > 


CONTENTS 


CHAPTER  I. 

DEFINITIONS  AND  FIRST  PRINCIPLES. 

CHAPTER  II. 

APPLICATIONS  AND  CONSTRUCTIONS. 

Page 

1.  To  find  the  Shadow  of  a Right  Line 16 

2.  To  find  the  Shadow  of  the  Abacus  and  Pillar 18 

3.  To  find  the  Shadows  of  the  Chimneys  and  House  . . . . 21 

4.  To  find  the  Shade  and  Shadow  of  the  Cylinder 25 

5.  To  find  the  Ellipse  from  its  Conjugate  Diameters  ....  30 

6.  To  find  the  Shadow  of  the  Rectangular  Abacus  on  a Column  32 

7.  To  find  the  Shadow  of  a Cylindrical  Abacus  on  a Column  . 33 

8.  To  find  the  Shadow  of  a Cylindrical  Abacus  on  a Wall  . . 34 

9.  To  find  the  Shadow  of  the  Inverted  Frustum  of  a Cone  . . 36 

10.  To  find  the  Shade  and  Shadow  of  a Sphere 40 

11.  Of  Brilliant  Points 44 

12.  To  find  the  Shade  and  Shadow  of  the  Ellipsoid  ....  46 

13.  To  find  the  Shadow  on  the  Interior  of  the  Niche  ....  49 

14.  To  find  the  Curve  of  Shade  on  the  Torus .55 

15.  To  find  the  Shadow  on  a Surface  of  Revolution  ....  57 

16.  To  find  the  Curve  of  Shade  on  the  Surface  of  Revolution  . 61 

17.  To  find  the  Line  separating  the  dark  from  the  illuminated 

part  of  a Surface  of  Revolution,  and  the  Shadow  cast  by 
the  Surface  ...*.' 66 

18.  To  find  the  Shades  and  Shadows  on  the  Base  and  Shaft  of 

the  Doric  Column 71 

19.  To  find  the  Shades  and  Shadows  on  the  Capital  of  the  Doric 

Column 74 

20.  Of  the  Helicoid 76 

21.  To  draw  a Tangent  Plane  to  the  Helicoid 79 

22.  To  find  the  Lines  of  Shade  and  Shadow  on  the  Screw  ...  81 


CONTENTS* 


VI  ij 


LINEAR  PERSPECTIVE 


CHAPTER  1. 

Paga 

1.  Definitions  and  First  Principles 95 

2.  To  find  the  Perspective  of  a Cube  and  the  Perspective  of  its 

Shadow 98 

CHAPTER  H. 

1.  Of  the  Method  of  Perspective  by  Diagonals  and  Perpendiculars  101 

2.  To  find  the  Perspective  of  a System  of  Cubes  and  the  Per- 


spective of  their  Shadows 107 

3.  To  find  the  Perspective  of  four  Pyramids 112 

4.  The  Perspectives  of  Tangent  Lines  are  tangent  to  each  other  116 

5.  To  find  the  Perspective  of  a Circle 117 

6.  To  find  the  Perspective  of  a Cylinder 119 

7.  To  find  the  Perspective  of  the  Frustum  of  an  Inverted  Cone  123 

8.  To  find  the  Perspective  of  a Niche 127 

9.  To  find  the  Perspective  of  a Sphere 131 

10.  To  find  the  Perspective  of  the  Groined  Arch 138 

11.  To  find  the  Perspective  of  a House  147 


SHADES  AND  SHADOWS. 


CHAPTER  I. 

DEFINITIONS  AND  FIRST  PRINCIPLES. 

1.  Light  is  that,  which  proceeding  from  an  object 
and  falling  upon  the  eye,  produces  the  sensation  of  sight. 

2.  Bodies  which  emit  or  give  out  light,  such  as  the 
sun,  a candle,  &c.,  are  called  luminous  bodies. 

3.  Light  is  supposed  to  emanate  from  every  pomt  of 
a luminous  body,  and  to  proceed  in  right  lines  when  not 
deflected  or  turned  from  its  course. 

4.  The  right  lines,  along  which  the  light  is  supposed 
to  move,  are  called  rays  of  light, 

5.  Since  light  flows  from  each  point  of  a luminous 
body  in  every  direction,  the  rays  of  light  flowing  from 
a single  point  form  a system  of  diverging  lines.  These 
rays  are  more  divergent  when  the  luminous  object  is 
near';  the  divergency  diminishes  when  the  distance  to 
the  luminous  object  is  increased,  and  becomes  nothing, 
or  the  rays  become  parallel,  when  that  distance  is  infinite. 

6.  The  sun,  the  chief  source  of  light,  is  so  far  dis- 
tant from  the  earth,  that  the  rays  of  light  coming  from 
it  may  be  regarded  as  parallel.  As  it  is  only  proposed, 
in  this  elementary  treatise,  to  find  the  shades  and 
shadows  of  objects  as  they  appear  in  nature,  the  rays  of 
light,  in  the  constructions  here  given,  will  be  considered 
as  parallel  with  each  other. 


10 


TREATISE  ON 


Having  considered  the  subject  under  this  point  of 
view,  it  will  be  easy  to  extend  the  principles  developed, 
to  embrace  the  cases  in  which  the  rays  of  light  are  con- 
vergent or  divergent. 

7.  In  the  graphic  constructions  which  follow,  it  be- 
comes necessary  to  represent  rays  of  light  by  right  lines. 
When  a line  is  assumed  and  called  a ray  of  light,  it  is 
to  be  understood  as  merely  indicating  the  direction  of 
the  light  as  it  falls  upon  the  body  whose  shade  or  shadow 
is  considered. 

8.  Bodies  or  objects  may  be  divided  into  three  classes  : 

1°.  Luminous  bodies,  or  those  which  give  out  light ; 

such  as  the  sun,  a candle,  &c. 

2°.  Opaque  bodies,  or  those  which  intercept  the  light ; 
such  as  wood,  stone,  iron,  &c. 

3°.  Transparent  bodies,  or  those  through  which  the 
light  passes  freely ; such  as  water,  glass,  &c. 

9.  Light  emanating  from  a luminous  object,  and 
falling  upon  an  opaque  b3dy,  is  received  by  that  part  of 
the  body  which  is  towards  the  source  of  light.  This 
portion  of  the  surface,  on  which  the  light  falls,  is  called 
the  illuminated  part  of  the  body.  That  portion  on  which 
the  light  does  not  fall,  is  called  the  shade  of  the  body ; 
and  the  line  which  separates  the  illuminated  part  from 
the  shade  is  called  the  line  of  shade. 

Let  C (PI.  1,  Fig.  1),  a point  in  the  plane  of  the  paper, 
be  the  centre  of  a sphere,  CB  its  radius,  and  AF,  also 
in  the  plane  of  the  paper,  a ray  of  light.  The  rays  of 
light  being  parallel,  and  falling  upon  the  sphere  in  the 
direction  AF,  may  be  regarded  as  forming  a cylinder, 
of  which  AC  is  the  axis.  The  surface  of  this  cylinder 
is  tangent  to  the  sphere  in  the  great  circle  of  which  the 
line  BE,  drawn  at  right  angles  to  AC,  is  the  projection. 
^(Davies’  Des.  Geom.  108.) 


SHADES  AND  SHADOWS.  1 I 

All  the  rays  of  light  which  lie  within  the  surface  of  this 
cylinder,  pierce  the  sphere  in  the  illuminated  part.  But 
the  sphere  being  supposed  opaque,  the  light  is  inter- 
cepted, and  does  not  fall  on  the  opposite  hemisphere. 
Hence, 

V.  The  surface  of  the  hemisphere  BFE,  towards  the 
source  of  light,  is  the  illuminated  part  of  the  sphere. 

2°.  The  surface  of  the  hemisphere  BPE,  opposite  the 
source  of  light,  is  the  shade  of  the  sphere. 

3°.  The  great  circle  of  which  BE  is  the  projection,  is 
the  line  of  shade. 

10.  The  shadow  of  an  object  is  that  part  of  space 
from  which  the  object  excludes  the  light  of  a luminous 
body.  Hence,  the  opaque  bodies  only  cast  shadows. 

Referring  again  to  the  sphere  of  which  C is  the  centre, 
and  to  the  cylinder  of  which  AC  is  the  axis,  it  follows, 
from  the  above  definition,  that  all  the  space  contained 
within  the  surface  of  the  cylinder,  estimated  from  BE, 
in  the  direction  CP,  is  but  the  indefinite  shadow  cast  by 
the  sphere. 

11.  Let  us  now  suppose  a body  to  be  placed  at  any 
distance  from  the  sphere,  and  opposite  to  the  source  of 
light.  It  is  plain  that  the  sphere  will  prevent  the  light 
from  falling  upon  a part  of  the  surface  of  the  body  so 
placed.  The  part  of  the  surface  from  which  the  light  is 
excluded  is  called  the  shadow  on  the  body  ; and  the  boun- 
dary of  this  shadow,  is  called  the  line  of  shadow. 

This  shadow  on  the  body  is  contained  within  the  sur- 
face of  the  cylinder  of  rays  that  is  tangent  to  the  sphere, 
and  the  line  of  shadow  is  the  line  of  intersection  of  the 
surface  of  this  cylinder  with  the  body  on  which  the 
shadow  falls. 

12.  If  a single  point,  in 'space,  be  supposed  to  inter- 
cept the  light,  the  direction  of  its  shadow  will  be  the 


12 


TREATISE  ON 


same  as  that  of  the  ray  passing  through  it;  and  the 
shadow  will  lie  on  the  opposite  side  of  the  point  from 
the  source  of  light. 

13.  If  the  light  be  intercepted  by  a right  line,  the 
shadow  will  be  determined  by  drawing  rays  of  light 
through  all  its  points.  These  rays  being  parallel,  and 
intersecting  the  same  right  line,  are  all  contained  in  a 
plane  passing  through  the  line  so  intersected.  This 
plane  is  called  a plane  of  rays, 

14.  The  shadow  of  a curved  line  is  determined  by 
drawing  rays  of  light  through  all  its  points.  These  rays, 
being  a system  of  parallel  lines,  may  be  regarded  as 
forming  the  surface  of  a cylinder  passing  through  the 
curve.  This  surface  is  called  the  surface  of  a cylinder 
of  rays, 

15.  Applying  what  has  been  shown  of  the  sphere  to 
any  opaque  body,  and  recapitulating,  in  part,  what  has 
already  been  said,  the  following  definitions  and  prin- 
ciples may  be  laid  down. 

1°.  The  illuminated  part  of  a body  is  that  portion  of 
its  surface  on  which  the  light  falls. 

2®.  The  shade  of  a body  is  that  part  of  the  surface 
from  which  the  light  is  excluded  by  the  body  itself. 

3°.  The  line  of  shade  is  the  line  separating  the  shade 
from  the  illuminated  part  of  a body,  and  is  the  line  of 
contact  of  a cylinder  of  rays  tangent  to  the  body. 

4°.  The  indefinite  shadow  of  a body  is  that  part  of 
space  from  which  the  body  excludes  the  light,  and  is 
limited  by  the  surface  of  the  tangent  cylinder  of  rays. 

5®.  The  shadow  on  a body  is  that  portion  of  its  sur- 
face from  which  the  light  is  excluded  by  an  opaque  body, 
between  it  and  the  source  of  light.  The  boundary  of 
this  shadow  is  the  line  of  shadow. 

6®.  The  line  of  shadow  is  the  intersection  of  the  sur- 


SHADES  AND  SHADOWS. 


13 


face  of  a cylinder  of  rays,  tangent  to  the  opaque  body, 
with  the  surface  on  which  the  shadow  falls. 

7®.  The  shadow  cast  by  a point  upon  any  surface,  is 
determined  by  finding  where  a ray  of  light  drawn 
through  the  point, pierces  the  surface. 

8®.  The  shadow  cast  by  a right  line  upon  any  surface 
is  the  intersection  of  a plane  of  rays  passing  through 
the  line  with  the  surface. 

9^  The  shadow  of  a curve  upon  any  surface  is  the 
intersection  of  the  surface  of  a cylinder  of  rays  passing 
through  the  curve,  with  the  surface  on  which  the  shadow 
falls. 

10".  The  line  of  shade  upon  a body  is  always  a 
line  of  contact,  and  the  line  of  shadow  a line  of  inter- 
section. 

11".  Since  the  line  of  shadow  is  determined  by  the 
surface  of  the  tangent  cylinder  of  rays,  it  follows,  that 
the  line  of  shade  is  the  line  which  casts  the  line  of 
shadow  on  the  body  where  the  shadow  falls. 

12".  When  the  curve  casting  the  shadow  is  a plane 
curve,  and  has  such  a position  that  its  plane  is  a plane 
of  rays,  the  rays  of  light  drawn  through  all  its  points 
will  form  a plane  and  not  a cylindrical  surface.  In  such 
case,  the  shadow  of  the  curve  upon  a surface  is  the  inter- 
section of  its  plane  with  the  surface. 

16.  If  a right  line  is  tangent  to  a*  curve  in  space,  the 
shadow  of  the  right  line  will  be  tangent  to  the  shadow 
of  the  curve.  F or,  the  plane  of  rays  which  determines 
the  shadow  of  the  right  line  will  be  tangent  to  the  cyl- 
inder of  rays  which  determines  the  shadow  of  the  curve ; 
therefore,  their  intersections  by  the  surface  on  which  the 
shadows  fall  are  tangent  to  each  other. 

17.  If  two  curves  are  tangent  to  each  other  in  space, 
their  shadows  on  any  surface  will  also  be  tangent. 


14 


TREATISE  ON 


F or,  the  cylindrical  surfaces  which  determine  the  shad- 
ows of  the  curves  are  tangent  to  each  other ; hence, 
the  curves  in  which  they  intersect  the  surface  on  which 
the  shadows  fall,  are  also  tangent  to  each  other. 

18.  The  shade  of  an  object  being  considerably 
darker  than  the  illuminated  part,  is  easily  distinguished 
from  it;  and  the  hne  of  shade  is,  in  general,  distinctly 
marked.  In  the  drawings,  the  shade  will  be  distinguished 
by  small  parallel  lines,  as  in  the  hemisphere  EPB  (PI. 
1,  Fig.  1). 

Shadows  also,  appearing  darker  than  that  part  of  the 
same  surface  on  which  the  shadow  does  not  fall,  will,  in 
the  drawings,  be  darkened  by  small  parallel  lines,  in  the 
same  manner  as  the  shade. 

19.  In  the  drawings  to  be  made,  two  planes  of  projec 
tion  will  be  used,  as  in  Descriptive  Geometry. 

20.  Since  the  projections  of  an  object  should  rep- 
resent the  appearance  which  the  object  itself  presents  to 
the  eye,  situated  at  an  infinite  distance  from,  and  in  a 
perpendicular  to,  the  plane  on  which  the  projection  is 
made,  only  that  part  of  its  surface  must  be  shaded  in 
the  drawing  which  is  in  the  shade,  and  seen  by  the  eye. 
The  portion  of  the  surface  which  is  in  the  shade  and  not 
seen  by  the  eye,  must  not  be  darkened  in  the  drawing ; 
if  it  were,  the  drawing  would  not  be  a true  representa- 
tion of  the  object. 

In  like  manner,  a shadow  upon  one  of  the  planes  of 
projection,  or  upon  any  other  surface,  may  be  concealed 
from  the  eye,  in  which  case  it  is  not  to  be  represented  in 
the  drawing. 

21.  The  same  general  rules  are  observed  in  making 
the  projections  of  bodies  as  in  Descriptive  Geometry. 
All  the  bounding  lines  which  are  seen  are  made  full ; 
and  all  auxiliary  lines  either  dotted  or  broken. 


SHADES  AND  SHADOWS. 


15 


The  rays  of  light  will  be  represented  by  small  broken 
lines. 

22.  To  enable  us  to  determine  the  shade  and  shadow 
of  an  object,  there  must  be  given, 

V.  The  position  of  the  opaque  object  casting  the 
shadow,  which  is  determined  when  its  projections  are 
given. 

2°.  The  surface  on  which  the  shadow  falls. 

3°.  The  direction  of  the  light. 

From  these  data,  the  shade  and  shadow  of  any  object 
can  always  be  found. 

The  terms  plan  and  elevation^  are  generally  used  by 
Architects  instead  of  the  terms,  horizontal  projection  and 
vertical  projection,  As’the  terms  are  synonymous,  either 
may  be  used. 

In  architectural  drawings  it  is  customary  to  suppose 
the  source  of  light  on  the  left  of  the  object,  and  the 
rays  to  have  such  a direction  that  their  horizontal  and 
vertical  projections  shall  make  angles  of  45°  with  the 
ground  line.  When  this  is  the  case,  it  is  plain  that  the 
rays  will  be  parallel  to  the  diagonal  of  a cube  whose 
faces  are  parallel  and  perpendicular  to  the  planes  of 
projection ; and  since  the  diagonal  of  a cube  makes  an 
angle  of  35°  16'  with  the  plane  of  either  of  its  faces, 
it  follows  that  any  ray  of  light  must  make  an  angle  of 
35°  16'  with  the  planes  of  projection,  when  both  its  pro 
jections  make  angles  of  45°  with  the  ground  line.  ^ 


16 


TREATISE  ON 


CHAPTER  11. 

APPLICATIONS  AND  CONSTRUCTIONS. 

PROBLEM  I. 

To  find  the  shadow  cast  by  a right  line  upon  the  horizontal 
plane  of  projection. 

23.  Let  IL  (PL  1,  Fig.  2)  be  the  ground  line,  (BD, 
B'D')  be  the  given  line ; (A, A)  a ray  of  light,  A being 
its  horizontal,  and  A'  its  vertical  projection. 

Through  any  point  of  the  line,  as  (B,  B'),  conceive  a 
ray  of  light  to  be  drawn.  Its  projections  will  be  respec- 
tively parallel  to  A and  A',  and  the  ray  will  pierce  the 
horizontal  plane  at  E,  which  will  be  one  point  in  the 
shadow  of  the  right  line. 

Through  any  other  point  of  the  right  line,  as  (D,  D') 
conceive  a ray  of  light  to  be  drawn.  Its  projections  will 
also  be  parallel,  respectively,  to  A and  A',  and  the  ray 
will  pierce  the  horizontal  plane  at  F,  which  is  a second 
point  in  the  shadow  of  the  given  line. 

Since  the  shadow  of  a right  line  upon  a plane  is  a 
right  line,  the  line  EF  is  the  shadow  cast  upon  the  hori- 
zontal plane  by  the  line  (BD,  B'D').  The  indefinite 
right  line  HG  is  the  indefinite  shadow  on  the  horizontal 
plane,  cast  by  an  indefinite  right  line  passing  through  the 
two  points  (B,B')  and  (D,D').  It  is  plain  that  this  shadow 
HG,  is  the  horizontal  trace  of  a plane  of  rays  passing 
through  the  line  (BD,  B'D'). 

The  right  line  TBD,  B'D)  may  be  so  situated  that  the 


SHADES  AND  SHADOWS. 


17 


whole,  or  a part  of  its  shadow,  will  fall  on  the  vertical 
plane.  If  it  be  required  to  find  that  shadow,  we  have 
only  to  construct  the  vertical  trace  of  the  plane  of  rays 
passing  through  the  given  line. 

It  IS  also  apparent,  that  the  point  H,  in  which  the  line 
(BD,  B'D')  pToduced,  pierces  the  horizontal  plane,  is  in 
the  right  line  joining  the  points  E and  F,  since  the  three 
points  are  all  in  the  trace  of  the  same  plane. 

Hence  we  conclude,  that  the  point  in  which  a right  line 
pierces  a plane  is  one  point  of  the  indefinite  shadow  of  the  line 
upon  the  plane.  And,  extending  the  principle,  the  indefinite 
shadow  cast  by  a right  line  upon  any  surface,,  passes  through 
the  point  in  which  the  right  line  ^ produced  if  necessary,,  pierces 
the  surface. 

24.  When  a right  line  is  parallel  to  the  plane  on  which 
the  shadow  falls,  the  shadow  will  be  parallel  to  the  line 
itself  For  if  the  line  and  its  shadow  be  not  parallel, 
they  would,  if  produced,  intersect,  which  they  cannot  do, 
since  a line  cannot  intersect  a plane  to  which  it  is  parallel. 

Construction  of  the  figure.  Draw  on  the  paper  the  inde- 
finite line  IL,  to  represent  the  intersection  of  the  planes 
of  projection,  or  the  ground  line.  Then  draw  the  lines 
BD,  B'D'  to  represent  the  projections  of  the  given  line ; 
also,  A and  A'  to  represent  the  projections  of  the  ray  of 
light.  This  being  done,  through  B and  D,  draw  lines 
parallel  to  A*  Through  B'  and  D'  draw  lines  parallel  to 
A.  At  the  points  E'  and  F',  erect  in  the  horizontal 
plane  perpendiculars  to  the  ground  line,  and  mark  the 
points  E and  F in  which  they  meet  the  parallels  first 

drawn.  Draw  the  line  EF,  which  is  the  shadow  required 

B 


18 


TREATISE  ON 


PROBLEM  II. 

Having  a rectangular  pillar  with  a rectangular  abacus 
placed  upon  its  upper  face  ; it  is  required  to  find  the  shadow 
cast  by  the  abacus  on  the  faces  of  the  pillar^  and  the  shadow 
cast  by  the  abacus  and  pillar  on  the  horizontal  plane. 

25.  Let  the  square  ah  (PI.  I.  Fig.  3)  be  the  horizontal 
projection  of  the  pillar,  cd  its  vertical  projection,  EC  the 
horizontal  projection  of  the  abacus,  and  E'G  its  vertical 
projection.  The  direction  of  the  light  is  shown  by  the 
ray  (A,  A'). 

Let  us  first  find  the  shadow  of  the  line  (BC,  FG),  on 
the  front  face  of  the  pillar.  Through  (B,  F)  a point 
of  the  line,  conceive  a ray  of  light  to  be  drawn ; the 
two  projections  are  respectively  parallel  to  A and  A'. 
The  point  in  which  this  ray  pierces  the  front  face 
of  the  pillar,  is  one  point  in  the  shadow  of  the  line  (BC, 
FG)  on  that  face.  But  the  line  (BC,  FG)  being  par- 
allel to  the  front  face,  the  shadow  is  parallel  to  the  line 
itself;  hence  A'm,  drawn  parallel  to  the  ground  line,  is 
the  vertical  projection  of  the  required  shadow.  The 
horizontal  projection  is  in  the  line  hb. 

The  line  of  which  BE  is  the'  horizontal  projection,, 
and  which  is  vertically  projected  at  F,  casts  a shadow 
upon  the  front  face  of  the  pillar,  on  the  face  wa,  and  on 
the  horizontal  plane. 

Conceive  a plane  of  rays  to  be  passed  through  this 
line — this  plane  is  perpendicular  to  the  vertical  plane 
of  projection,  and  its  vertical  trace  Yn'h\  is  parallel  to 
the  vertical  projection  of  the  light;  and  nU  is  also  the 
projection  of  its  intersection  with  the  front  face  of  the 
pillar : therefore  nU‘  is  the  vertical  projection  of  the 


SHADES  AND  SHADOWS. 


19 


shadow  cast  on  the  front  face  of  the  pillar  by  a part  of 
the  line  (BE,  F) — the  horizontal  projection  of  the  same 
shadow  is  nh.  Drawing  through  w,the  line  pn  parallel 
to  the  horizontal  projection  of  the  light,  gives  Bp  the 
horizontal  projection  of  that  part  of  the  line  which 
casts  a shadow  on  the  front  face. 

The  part  of  the  line  from  p to  casts  a right  line  of 
shadow  on  the  face  na^  which  shadow  is  vertically  pro- 
jected at  n.  The  part  casts  a shadow  on  the  horizon- 
tal plane.  In  finding  the  shadow  of  the  abacus  on  the 
horizontal  plane,  we  will  begin  with  the  point  (C,  D'). 

Through  this  point  suppose  a ray  of  light  to  be  drawn. 
Such  ray  will  pierce  the  horizontal  plane  at  H.  Through 
(D,  D')  conceive  a ray  also  to  be  drawn.  Such  ray  will 
pierce  the  horizontal  plane  at  I.  Hence  HI  is  the 
shadow  cast  upon  the  horizontal  plane  by  the  line 
(CD,D). 

But  I is  also  the  shadow  on  the  horizontal  plane  of 
one  point  of  the  line  (ED,  E'D') ; and  since  this  line  is 
parallel  to  the  plane,  IL  drawn  parallel  to  ED,  is  its 
indefinite  shadow.  The  shadow  is  limited  by  the  line 
EL,  drawn  through  E,  parallel  to  the  horizontal  pro- 
jection of  the  light. 

Through  the-  point  (C,  G)  conceive  a ray  of  light  to 
be  drawn.  It  pierces  the  horizontal  plane  at  ^.  The 
line  is  the  shadow  cast  by  the  vertical  line  (C,  GD') 
of  the  abacus. 

Through  the  point  ^,draw  iK  parallel  to  BC,  and  pro- 
duce it  till  it  meets  ^K,  drawn  through  parallel  to  the 
horizontal  projection  of  the  light.  Then  Ki  is  the  shadow 
cast  by  (^C,  A;'G),a  part  of  the  front  and  lower  line  of 
the  abacus.  The  ray  of  light  through  the  point  (k,  k') 
touches  the  edge  of  the  pillar  at  m,  and  pierces  the  hori- 
zontal plane  at  K. 


20 


TREATISE  ON 


From  K to  6 the  shadow  is  cast  by  the  part  of  the 
edge  {b^mdy 

Returning  to  the  point  L,  L/  is  the  shadow  cast  by  the 
line  (E,  E'F),y^  the  shadow  cast  by  the  line  (Egr,  F),  and 
ga  the  shadow  cast  by  that  part  of  the  edge  between  n 
and  the  horizontal  plane. 

The  light  falls  on  three  faces  of  the  abacus,  viz : the 
upper  face,  whose  vertical  projection  is  the  line  E'D'; 
the  side  face  whose  vertical  projection  is  the  line  E'F ; 
and  the  front  face  whose  vertical  projection  is  the  rec-^ 
tangle  FD'.  The  three  remaining  faces  are  in  the  shade, 
but  neither  of  them  is  seen  in  either  projection. 

The  part  of  the  front  face  of  the  pillar  on  which  the 
shadow  of  the  abacus  falls,  and  the  shadow  of  the  pil- 
lar and  abacus  on  the  horizontal  plane,  are  shaded. 
The  part  of  the  shadow  on  the  horizontal  plane  which 
falls  under  the  abacus  cannot  be  seen  in  the  horizontal 
projection,  and  is  therefore  not  shaded. 

It  may  not  be  out  of  place  to  remark,  that  the  lines 
of  shadow  which  have  been  determined,  are  but  the  in- 
tersections of  planes  of  rays  passing  through  the  lines 
casting  the  shadows,  with  those  planes  on  which  the 
shadows  fall. 

Thus  Km  is  the  trace,  on  the  front  face  of  the  pillar, 
of  a plane  of  rays  passing  through  (BC,  FG).  The 
shadow  nK  is  the  trace,  on  the  front  face  of  the  pillar, 
of  a plane  of  rays  passing  through  the  line  (EB,  F) ; HI 
is  the  trace,  on  the  horizontal  plane,  of  a plane  of  rays 
passing  through  (CD,  D') ; Hi  is  in  the  horizontal  trace  of 
a vertical  plane  of  rays  passing  through  the  vertical  line 
(C,  D'G).  The  shadow  ^K  is  in  the  horizontal  trace  of 
a plane  of  rays  passing  through  (BC,  FG).  The  shadow 
K6  is  in  the  horizontal  trace  of  the  vertical  plane  of 
]*ays  passing  through  the  edge  (^,  dm^  ; and  this  plane 


SHADES  AND  SHADOWS. 


21 


cuts  off  the  part  (^C,  ^'G)  of  the  line  (BC,  FG),  which 
casts  the  shadow  Ke  on  the  horizontal  plane.  Similar 
explanations  may  be  given  of  the  other  lines  of  shadow 
on  the  horizontal  plane. 

PROBLEM  III. 

It  is  required  to  find  the  shadows  cast  hy  the  cornices  of 
the  chimneys  of  a house^  on  the  faces  of  the  chimneys ; the 
shadoivs  cast  by  the  cornices  and  chimneys  on  the  roof  of  the 
house^  and  the  shadows  cast  hy  the  roof  on  the  walls ^ and  on 
a horizontal  plane  at  a given  distance  below  the  eaves, 

26.  Let  the  rectangle  BCEA  (PI.  2)  be  the  horizontal 
projection  of  the  outel*  lines  of  the  eave-trough.  These 
lines  are  contained  in  a horizontal  plane,  and  are- ver- 
tically projected  in  the  line  A'C'.  The  lines  of  the  inner 
rectangle  are  the  intersections  of  the  outer  faces  of 
the  walls  with  the  horizontal  plane;  they  are  made 
broken  in  projection,  being  concealed  by  the  roof.  The 
lines  of  the  middle  rectangle  are  the  horizontal  pro- 
jections of  the  eaves  of  the  roof.  The  eaves  are 
supposed  to  be  in  the  same  horizontal  plane  with  the 
upper  line  of  the  eave-trough,  and  are  therefore  verti- 
cally projected  in  the  line  AC'. 

The  line  in  which  the  side  roofs  intersect  each  other, 
and  the  lines  in  which  the  side  roofs  intersect  the  end 
roofs,  are  represented  by  full  lines  in  horizontal  pro- 
jection. 

With  regard  to  the  chimneys,  the  lines  of  the  outer 
rectangle,  as  ifhq^  are  the  horizontal  projections  of  the 
outer  lines  of  the  cornice ; the  lines  of  the  middle 
rectangle  are  the  intersections  of  the  outer  faces  of  the 


22 


TREATISE  ON 


chimneys  with  the  roofs ; and  the  inner  rectangle  is  the 
projection  of  the  flue  or  hollow  part  of  the  chimney. 

The  vertical  projections  of  the  roofs  and  chimneys  can 
be  understood  from  the  figure,  excepting,  perhaps,  the 
manner  of  determining  the  lines  in  which  the  front  and 
back  faces  of  the  chimney  in  the  front  roof,  intersect  the 
roof.  After  the  horizontal  projection  of  the  chimney  is 
made,  these  lines  are  thus  found : 

The  faces  of  the  chimney  intersect  the  front  roof  in 
lines  parallel  to  the  ground  line.  Producing  the  hori- 
zontal projection  of  the  front  face  till  it  meets  in  G,  the 
line  of  the  end  roof,  we  obtain  IG,  the  horizontal  pro- 
jection of  the  line  of  intersection.  But  G is  vertically 
projected  in  G',  therefore  Gl'  drawn  through  G'  parallel 
to  the  ground  line,  is  the  vertical  projection  of  the  line 
in  which  the  front  face  of  the  chimney  intersects  the 
roof.  For  similar  reasons,  the  line  drawn  through  H' 
parallel  to  the  ground  line,  is  the  vertical  projection  of 
the  line  in  which  the  back  face  of  the  chimney  intersects 
the  roof. 

In  finding  the  shadows,  let  us  begin  with  the  chimney 
of  the  end  roof. 

Through  (/, /"),  a point  of  the  lower  line  of  the  cor- 
nice, conceive  a ray  of  light  to  be  drawn.  It  pierces  the 
front  face  of  the  chimney  in  the  point  {gig) ; and  the  line 
gp  is  the  vertical  projection  of  the  shadow  cast  by  (/A, 
f'W)  on  the  front  face  of  the  chimney.  The  line  og  is  the 
vertical  projection  of  the  shadow  cast  on  the  front  face 
by  a part  of  the  line  ( H.f").  A similar  construction  gives 
the^  shadow  cast  on  the  front  face  of  the  other  chimney., 

To  find  the  shadows  which  the  chimneys  cast  on  the 
roofs. 

Through  the  line  h')  conceive  a plane  of  rays  to 
be  passed.  Since  the  line  (hq,  H)  is  perpendicular  to 


SHADES  AND  SHADOWS. 


2.3 


the  vertical  plane  of  projection,  the  plane  of  rays  through 
it  will  also  be  perpendicular  to  the  vertical  plane ; and 
hence  its  vertical  trace  will  be  the  line  /iTJP  ’,  parallel  to 
the  vertical  projection  of  the  light.  This  plane  of  rays  in- 
tersects the  horizontal  plane  of  the  eaves  in  a line  which 
is  perpendicular  to  the  vertical  plane  at  F",  and  of  which 
F'F,  perpendicular  to  the  ground  line,  is  the  horizontal 
projection.  It  also  cuts  the  line  of  intersection  of  the 
side  roofs  in  a point  whose  vertical  projection  is  k'  and 
whose  horizontal  projection  is  k ; and  hence  kF^  kF'  are 
the  horizontal  projections  of  the  intersections  of  the 
plane  of  rays  through  h)  with  the  side  roofs ; which 
intersections  are  the  indefinite  shadows  cast  by  {hq^  K) 
on  these  roofs.  The  shadows  are  limited  in  horizontal 
projection  by  the  lines  hn  and  drawn  parallel  to  the 
horizontal  projection  of  the  light.  Therefore  kn  and 
km  are  the  horizontal  projections  of  the  shadows  cast 
on  the  side  roofs  by  the  line  Qiq^  h'). 

It  may  be  here  remarked,  that  all  planes  of  rays  which  are 
perpendicular  to  the  vertical  plane  of  projection  are  parallel 
to  each  other ^ and  will  consequently  intersect  the  side  roofs  in 
lines  parallel  to  kF'  and  kF, 

Drawing  through  m the  line  ms,,  parallel  to  the  ground 
line,  and  through  ^,  is  parallel  to  the  horizontal  projec- 
tion of  the  light,  gives  ms  for  the  shadow  of  the  line 
(iq-,fk)  on  the  side  roof.  The  line  us, is  the  horizontal 
projection  of  the  shadow  cast  by  the  perpendicular  (i, 
/'/").  This  shadow  is  in  the  trace  of  a vertical  plane 
of  rays  through  {hff”),  and  is  limited  at  u by  the  intersec- 
tion of  a plane  of  rays  through  (/^/")  with  the  side  roof. 

The  line  uv,  lying  in  this  intersection,  is  the  shadow 
cast  by  a part  of  the  line  (fhfj  on  the  side  roof  The  • 
shadow  at  v is  cast  by  that  point  of  the  back  edge  of  the 
chimney  which  is  vertically  projected  at  o. 


24 


TREATISE  ON 


From  V the  shadow  is  cast  by  the  back  edge  of  the 
chimney. 

Returning  to  the  shadow  cast  by  the  point  {h^  A'),  we 
find,  first,  the  shadow  nx^  which  is  cast  by  the  perpen- 
dicular (A,  A' A"),  and  then  draw  through  a;_a  .parallel  to 
the  ground  line ; this  parallel  is  the  indefinite  shadow 
cast  by  the  line  (/A, /"A")  on  the  side  roof  We  next 
find  the  point  in  which  a ray  of  light  through  the  point 
whose  vertical  projection  is  pierces  the  end  roof, 
which  is  at  (/,  /).  Joining  the  point  t with  the  point  in 
which  the  parallel  through  x meets  the  intersection  of 
the  side  and  end  roofs,  gives  the  shadow  cast  by  the  line 
on  the  end  roof  From  t th^  shadow  is  cast 
by  the  edge  of  the  chimney. 

We  find  the  vertical  projection  of  this  shadow  by  pro- 
jecting its  points  into  the  vertical  traces  of  the  planes 
to  which  they  respectively  belong.  Thus  m is  verti- 
cally projected  at  m',  s at  s\  u at  u\  v at  y at  y\  k at 
k\  n at  w , and  x at  x.  That  part  of  the  shadow  which 
is  on  the  front  roof  is  all  that  can  be  seen  in  vertical 
projection.  The  boundary  of  that  on  the  back  roof 
is  therefore  dotted. 

The  shadows  of  the  other  chimney  are  found  in  a 
manner  so  entirely  similar  as  not  to  require  a particular 
explanation. 

Let  us  n ow  find  the  shadows  on  the  walls  of  the  house. 

Through  (BC,  ac)^  the  upper  line  of  the  convex  part 
of  the  eave-trough,  conceive  a plane  of  rays  to  be  passed. 
Its  trace  on  the  front  wall  limits  the  shadow  cast  by  the 
eave-trough.  The  point  (5,  b')  is  in  the  trace  of  this 
plane,  and  He  is  the  required  line  of  shadow*  The  shadow 
hz  is  cast  by  the  corresponding  line  of  the  end  eave- 
trough,  which  also  casts  a shadow  on  the  end  wall  that  is 
vertically  projected  at  r. 


SHADES  AND  SHADOWS. 


25 


To  find  the  shadow  cast  by  the  house  on  the  horizontal 
plane  of  projection. 

Through  (CE,  C'),  the  upper  line  of  the  end  eave- 
trough,  conceive  a plane  of  rays  to  be  passed.  The  line 
LE'  is  its  trace  on  the  horizontal  plane,  and  consequently 
the  shadow  of  the  line  (CE,  C'). 

The  small  vertical  line  (C,  cC')  casts  its  shadow  Lc 
in  the  line  CL.  Through  c draw  cd  parallel  to  the 
ground  line,  and  produce  it  till  it  meets  Dc?  drawn  par- 
allel to  the  horizontal  projection  of  the  light.  Then 
dc  is  the  shadow  cast  by  the  line  (DC,D'c),  and  the  ray 
through  (D,  D')  touches  the  corner  of  the  house  at  e, 
and  pierces  the  horizontal  plane  at  d. 

The  line  a E',  drawn  through  E'  parallel  to  the  ground 
line,  and  limited  by  Kd  drawn  parallel  to  the  horizontal 
projection_of  the  light,  is  the  shadow  cast  on  the  hori- 
zontal plane  by  the  line  (AE,  A'C').  The  line  (A,  a A') 
- casts  the  shadow  dr ; a part  of  the  line  (BA,  a)  casts  the 
shadow  rr  ; from  r the  line  of  shadow  is  cast  by  the 
corner  of  the  house  below  2^. 

The  shadow  on  the  horizontal  plane  which  lies  under 
the  eave-trough  and  cornice,  is  not  seen  in  horizontal 
projection. 

PROBLEM  IV. 

Havi7ig  gwen  an  oblique  cylinder^  and  the  direction  of  the 
lights  it  is  required  to  find  the  shade  on  the  exterior  surface^ 
the  shadow  of  the  upper  circle  on  the  interior  surface^  and  the 
shadow  of  the  cylinder  on  the  horizontal  plane, 

27.  Let  (AB,  AB'),  (PI.  3.  Fig.  1),  be  the  axis  of  the 
cylinder,  and  suppose  the  projections  to  be  made  as 
seen  in  the  figure. 


26 


TREATISE  ON 


If  we  suppose  two  tangent  planes  of  rays  to  be  drawn 
to  the  surface  of  the  cylinder,  the  elements  of  contact 
will  separate  the  dark  from  the  illuminated  part  of  the 
surface,  and  will  consequently  be  lines  of  shade. 

In  order  to  pass  these  planes,  we  first  pass  a plane  of 
rays  through  the  axis  of  the  cylinder,  to  which  the 
tangent  planes  will  both  be  parallel. 

• Through  (B,B'),  a point  of  the  axis,  conceive  a ray  of 
light  to  be  drawn — it  pierces  the  horizontal  plane  at  C. 
The  axis  of  the  cylinder  pierces  the  horizontal  plane  at 
A — therefore  AC  is  the  horizontal  trace  of  a plane  of . 
rays  passing  through  the  axis  of  the  cylinder. 

Let  two  lines  be  now  drawn  tangent  to  the  base  of  the 
cylinder,  and  parallel  to  AC.  These  lines  are  De  and 
G/'',  and  are  also  the  horizontal  traces  of  the  planes 
of  rays  tangent  to  the  surface  of  the  cylinder.  The 
elements  of  contact  pierce  the  horizontal  plane  at  D and 
G.  Hence,  DE  and  GF,  drawn  parallel  to  AB,  are  the 
horizontal  projections  of  the  elements  of  contact : and 
by  projecting  D and  G into  the  ground  line  at  D'  and  G', 
and  E and  F into  the  vertical  projection  of  the  upper 
circle  at  E'  and  F',  we  determine  D'E'  and  G'F',  the 
vertical  projections  of  the  elements  of  contact.  The 
points  D,  A and  G,  are  in  the  same  straight  line  perpen- 
dicular to  AC ; and  DHG  is  a semicircle. 

The  light  falls  on  the  half  of  the  exterior  surface  cor- 
responding to  the  semicircle  DHG — the  remaining  half 
is  in  the  shade.  The  lines  (DE,  D'E',)  (GF,  G'F')  are 
lines  of  shade,  and  being  also  elements  of  the  surface, 
are  generally  called  elements  of  shade. 

In  the  horizontal  projection,  we  do  not  see  the  shade 
which  is  on  the  under  surface  of  the  cylinder— therefore, 
we  darken  only  that  part  which  is  included  between  the 
elements  DE  and  LN.  In  the  vertical  projection,  we 


SHADES  AND  SHADOWS. 


27 


see  that  part  of  the  shade  included  between  the  element 
GT',  and  the  element  P'K',  only  that  portion  of  the 
surface,  therefore,  is  shaded. 

To  find  the  shadow  cast  by  the  upper  circle  on  the 
interior  surface. 

This  shadow  begins  at  the  points  (E,  E'),  (F,F')  in 
which  the  elements  of  shade  meet  the  upper  circle,  and 
is  cast  by  the  semicircle  whose  horizontal  projection  is 
FdaE. 

If  the  surface  be  intersected  by  planes  of  rays  parallel 
to  the  axis  of  the  cylinder,  each  plane  so  drawn  will  in- 
tersect the  surface  in  two  rectilinear  elements.  If  then, 
through  the  upper  extremity  of  the  element  towards  the 
source  of  light,  a ray  be  drawn,  it  will  be  contained  in 
the  plane  of  rays,  and  will,  consequently,. intersect  the 
other  element  lying  in  that  plane — the  point  of  inter- 
section is  a point  of  shadow  on  the  interior  surface. 
But  the  horizontal  traces  of  all  planes  of  rays  parallel 
to  the  axis  of  the  cylinder  are  parallel  to  AC,  the 
trace  of  the  plane  of  rays  through  the  axis.  Therefore, 
draw  any  line,  as  IP,  parallel  to  AC,  to  represent  the 
trace  of  a plane  of  rays  parallel  to  the  axis.  This  plane 
intersects  the  surface  in  two  elements,  whose  horizontal 
projections  are  Id  and  PK.  Through  J,  the  upper  ex- 
tremity of  the  element  towards  the  source  of  light, 
draw  df  parallel  to  the  horizontal  projection  of  the  light 
-the  point /,  in  which  it  meets  PK,  is  the  horizontal  pro- 
jection of  a point  of  shadow  on  the  inner  surface. 
The  vertical  projection  of  this  point  is  found  by  pro- 
jecting P into  the  ground  line,  drawing  the  vertical  pro- 
jection of  the  element  on  which  the  shadow  falls,  and 
noting  its  intersection  with  the  perpendicular  to  the 
ground  line,  drawn  through  the  point  /;  or,  by  pro- 
jecting the  point  d into  the  upper  base,  drawing  through 


28 


TREATISE  ON 


d the  vertical  projection  of  a ray  of  light,  and  noting 
its  intersection  /'with  the  vertical  projection  of  the 
element  on  which  the  shadow  falls. 

If  it  be  required  to  find  the  shadow  on  any  particular 
element,  as  the  one,  for  example,  whose  horizontal  pro- 
jection is  LN,  we  have  merely  to  pass  a plane  of  rays 
through  the  element,  and  determine  the  other  element 
towards  the  source  of  light,  in  which  it  intersects  the 
surface.  Then,  through  the  upper  extremity  of  the 
element  casting  the  shadow,  draw  a ray  of  light,  and 
where  it  meets  the  given  element,  is  the  required  point 
of  the  shadow.  If  we  take  the  element  whose  hori- 
zontal projection  is  LN,  RL  is  the  horizontal  trace  of 
the  plane  of  rays  passing  through  it — the  element 
towards  the  source  of  light,  in  which  this  plane  inter- 
sects the  surface,  is  horizontally  projected  in  R^,  and  g 
is  the  horizontal  projection  of  the  point  of  shadow  cast 
by  its  upper  extremity. 

At  the  horizontal  projection  of  the  curve  of 
shadow  is  tangent  to  the  line  LN : for  the  vertical  plane 
of  which  LN  is  the  horizontal  trace,  is  tangent  to  the 
vertical  cylinder,  which  projects  the  curve  of  shadow, 
along  the  element  that  pierces  the  horizontal  plane  at^. 

The  lowest  point  of  the  curve  of  shadow,  or  that 
point  which  is  farthest  from  the  upper  circle,  is  found  by 
passing  a plane  of  rays  through  the  axis.  F or,  since  all 
the  rays  of  light  make  equal  angles  with  the  horizontal 
plane,  or  with  the  plane  of  the  upper  base,  the  distance 
of  the  points  of  shadow  below  the  upper  base  will 
increase,  as  the  distance  between  the  element  cashing 
and  the  element  receiving  the  shadow,  is  increased.  But, 
the  plane  of  rays  through  the  axis  intersects  the  surface 
m elements  farther  from  each  other  than  the  elements  cut 
put  by  any  other  of  the  parallel  planes  of  rays.  Hence, 


SHADES  AND  SHADOWS. 


29 


it  determines  the  lowest  point  of  shadow,  which  is 

Having  found  as  many  points  of  the  shadow  as  may 
be  necessary,  let  the  projections  of  the  curve  be  accu- 
rately described. 

In  horizontal  projection,  the  curve  of  shadow  is  seen 
from,  F until  it  crosses  the  horizontal  projection  of  the 
upper  circle,  thence  to  the  point  E it  is  concealed  by 
the  surface  of  the  cylinder.  The  part  of  the  inner  sur- 
face lying  above  the  curve  of  shadow,  receives  the  light ; 
the  part  lying  below  it  is  in  the  shadow.  That  part  of 
the  surface  which  is  in  the  shadow,  and  seen,  is  shaded 
in  the  drawing.  With  respect  to  the  vertical  projection, 
no  part  of  the  curve  of  shadow  is  seen,  since  it  lies  en- 
tirely on  the  inner  surface  of  the  cylinder. 

To  find  the  shadow  of  the  cylinder  on  the  horizontal 
plane. 

The  lines  of  the  cylinder  which  cast  shadows  on  the 
horizontal  plane  are,  the  two  elements  of  shade,  and 
the  half  of  the  upper  circle  which  is  opposite  the  source 
of  light,  and  of  which  ENF  is  the  horizontal  pro- 
jection. 

The  shadows  cast  by  the  elements  of  shade  fall  in  the 
horizontal  traces  of  the  tangent  planes  of  rays — hence, 
we  may  consider  De  and  G/",  as  the  indefinite  shadows 
pf  the  elements  of  shade. 

If  through  the  upper  circle  of  the  cylinder  we  con- 
ceive a cylinder  of  rays  to  be  passed,  its  axis  will  be 
a ray  of  light  passing  through  the  centre  (B,  B'),  and 
will  pierce  the  horizontal  plane  at  C.  But,  since  the 
plane  of  the  upper  circle  and  the  horizontal  plane  are 
parallel,  they  will  intersect  the  surface  of  the  cylinder 
of  rays  in  equal  circles.  Therefore,  a circle  described 
with  C as  a centre,  and  a radius  equal  to  B'K',  will  be  the 


30 


TREATISE  ON 


shadow  cast  by  the  upper  circle  on  the  horizontal  plane. 
The  shadows  cast  by  the  elements  of  shade  will  be 
tangent  to  this  circle  of  shadow  at  the  points  f"  and  e ; 
which  points  are  found  by  drawing  a line  through  C per- 
pendicular to  AC,  or  by  drawing  the  horizontal  projec- 
tions of  rays  through  the  points  F and  E. 

Although  we  have  spoken  of  the  shadow  cast  on  the 
horizontal  plane  by  the  whole  of  the  upper  circle,  yet  it 
is  obvious  that  the  semicircle  which  is  towards  the  source 
of  light,  and  which  casts  a shadow  on  the  inner  surface 
of  the  cylinder,  cannot  cast  a shadow  on  the  hori- 
zontal plane,  unless  we  suppose  the  surface  to  be  trans- 
parent, so  that  the  light  may  not  be  intercepted  by  it. 

The  part  of  the  shadow  on  the  horizontal  plane 
which  is  concealed  by  the  horizontal  projection  of  the 
surface,  is  not  shaded  in  the  drawing. 

28.  The  shadow  of  a circle,  or  indeed  of  any  curve, 
on  a plane  tq  which  it  is  parallel,  is  an  equal  circle  or 
curve.  But  the  shadow  of  a circle  on  a plane  to  which 
it  is  not  parallel,  is,  in  general,  an  ellipse.  The  shadow 
cast  by  the  centre  of  the  circle  is  the  centre  of  the  ellipse, 
and  the  shadow  cast  by  any  diameter  of  the  circle  is  a 
diameter  of  the  ellipse.  Two  diameters  of  an  ellipse 
are  said  to  be  conjugate  when  either  of  them  is  parallel 
to  tangent  lines  drawn  through  the  vertices  of  the  other. 
If  any  two  diameters  be  taken  in  a circle  at  right  angles  to 
each  other ^ their  shadows  will  be  conjugate  diameters  of  the 
ellipse  of  shadow. 

For,  if  through  the  vertices  of  either  diameter,  tangent 
lines  be  drawn  to  the  circle,  they  will  be  parallel  to  the 
other  diameter ; hence,  their  shadows  on  any  plane  will 
be  parallel  to  the  shadow  of  this  latter  diameter.  But 
the  shadows  of  the  tangents  will  be  tangent  to  the 
shadow  of  the  circle  (16),  that  is,  to  the  ellipse  of 


SHADES  AND  SHADOWS. 


31 


shadow.  Therefore,  the  shadow  of  one  diameter  is 
parallel  to  the  tangents  drawn  through  the  vertices  of 
the  shadow  of  the  other.  Hence  the  shadows  cast  by 
two  diameters  of  a circle  at  right  angles  to  each  other, 
are  conjugate  diameters  of  the  ellipse  of  shadow. 

29.  It  is  often  required  to  construct  ellipses  when  their 
conjugate  diameters  only  are  known.  The  construction 
is  easiest  made  by  finding  the  axes  of  the  curve. 

We  shall,  therefore,  give  a problem,  the  principles  of 
which  are  found  inCrozet’s  Conic  Sections,  for  finding  the 
axes  of  an  ellipse  when  two  of  its  conjugate  diameters  are  given. 

Let  AB  and  DS  (PI.  3,  Fig.  2),  be  two  conjugate  diame- 
ters of  an  ellipse.  Through  either  vertex,  as  D,  of  either 
diameter,  draw  a parallel  EF  to  the  other  diameter.  At 
D erect  in  a plane  perpendicular  to  AOD,  DO'  perpen- 
, dicular  to  EF,  and  make  it  equal  to  half  the  parallel 
diameter  AB.  With  O'  as  a centre,  and  radius  O'D, 
describe  the  circle  CaDb. 

Now  we  may  regard  the  ellipse  whose  conjugate  di- 
ameters are  AB  and  DS,  as  the  shadow  of  the  circle 
Cam. 

The  ray  of  light  through  the  centre  O'  pierces  the 
horizontal  plane  at  O ; the  diameter  ah  casts  the  shadow 
AB,  and  the  diameter  CD  the  shadow  DS. 

It  is  now  required  to  find  two  diameters  of  the  circle 
CdDh  at  right  angles  to  each  other,  whose  shadows 
shall  also  be  at  right  angles;  for,  the  conjugate 
diameters  of  an  ellipse,  which  are  perpendicular  to  each 
other,  are  the  axes. 

To  find  these  diameters  it  is  necessary  to  construct 
two  semicircles  having  a common  diameter  in  the  line 
EF,  one  lying  in  the  horizontal  plane  and  passing 
through  O,  the  other  in  the  vertical  plane  and  passing 
through  O'. 


32 


TREATISE  ON 


Produce  O'D  and  make  DP  equal  to  it.  Join  OP  and 
bisect  it  by  a perpendicular  line — the  point  N,  where  the 
perpendicular  meets  EF,  is  the  common  centre  of  the 
two  semicircles.  Let  them  be  described  with  the  radius 
NO,  or  NO'.  Then  draw  the  radii  O'^E  and  O'/F. 
These  radii  are  at  right  angles  to  each  other ; and  so 
are  their  shadows  EO  and  FO.  Hence,  GO  and  LO  are 
the  semi-axes  of  the  ellipse.  The  extremities  G and  L 
are  determined  by  drawing  rays  through  g and  /,  the  ex- 
tremities of  the  perpendicular  radii. 

PROBLEM  V. 

To  find  the  shadow  of  a rectangular  abacus  on  a cylin^ 
drical  column  and  also^  the  shade  of  the  column, 

30.  Let  the  semicircle  zgp  (PL  3,  Fig.  3)  be  the  plan 
of  a semi-column,  and  opdc  its  elevation ; the  rectangle 
cp  the  plan  of  the  abacus,  and  c'd  its  elevation. 

Through  the  lower  line  (czf^  of  the  abacus,  conceive 
a plane  of  rays  to  be  passed.  This  plane  will  be  per- 
pendicular to  the  vertical  plane  of  projection,  and  its 
intersection  with  the  surface  of  the  column  will  be  the 
shadow  cast  by  the  line  (cz,,  cf  (15). 

Through  c draw  ci'  parallel  to  the  vertical  pro- 
jection of  the  light.  This  line  is  the  vertical  trace  of 
the  plane  of  rays  passed  through  the  line  (cz^  c'),  and  it 
is  also  the  vertical  projection  of  the  indefinite  shadow 
cast  on  the  column.  (Des.  Geom.  82.) 

But  the  ray  of  light  through  the  point  (c,  c)  pierces 
tlie  surface  of  the  column  at  (e,  e'),  therefore  ct  is  the 
vertical  projection  of  the  shadow  cast  on  the  column  by 
the  line  (cz,  c).  The  shadow  itself  is  an  ellipse,  and  is 
horizontally  projected  in  the  arc  zi. 


SHADES  AND  SHADOWS. 


33 


To  find  the  shadow  cast  by  the  line  (cc/,  co?')  : 

The  shadow  cast  by  the  point  (c,c')  is  already  found 
at  {i,iy 

Let  the  surface  of  the  column  be  intersected  by 
planes  of  rays  perpendicular  to  the  horizontal  plane. 
These  planes  will  cut  the  line  {cd^  cd'^  of  the  abacus  in 
points,  and  the  surface  of  the  column  in  right  lines— 
then,  drawing  through  the  points  of  the  abacus,  rays  of 
light,  where  they  meet  the  elements  of  the  column  are 
points  of  the  shadow. 

The  line  bk^  drawn  parallel  to  the  horizontal  projection 
of  the  light,  is  the  horizontal  trace  of  a vertical  plane 
of  rays,  and  this  plane  determines  (/c,  k'^  a point  of  the 
shadow.  The  points  (/,  (g, g-')  (^,  and  (w,  n)  are 

found  in  a similar  manner. 

The  vertical  plane  of  rays,  whose  horizontal  trace  is 
EF,  is  tangent  to  the  column.  The  ray,  therefore, 
through  the  point  (E,  E',)  touches. the  column  at  (w,  n). 
At  this  point  the  line  of  shadow  terminates,  and  the  line 
of  shade  begins  and  passes  down  the  column  vertically, 
being  the  element  in  which  the  plane  EF  is  tangent  to 
the  column. 

The  part  of  the  surface  of  the  column  which  is  aboye 
the  line  of  shadow,  and  the  portion  \vhich  is  in  the 
shade,  are  shade  in  the  drawing. 

PROBLEM  VI. 

To  find  the  shadow  cast  by  a cylindrical  abacus  on  a cylin- 
drical  column  ; and  ihe  shade  on  the  abacus  and  column, 

31.  Let  nsr  (PI.  6,  Fig.  2)  be  the  plan  of  a semi- 
column,// its  elevation — cmd  the  plan  of  the  abacus, 
and  cb  its  elevation. 

Let  the  surface  of  the  column  be  intersected  by  ver- 

C 


34 


TREATISE  ON 


tical  planes  of  rays.  These  planes  will  intersect  the 
lower  circle  of  the  abacus,  which  casts  the  line  of 
shadow  on  the  column,  in  points,  and  the  surface  of  the 
column  in  right  lines.  Through  the  points  of  the 
abacus  let  rays  of  light  be  drawn — the  points  in 
which  they  intersect  the  elements  of  the  column  are 
points  of  the  line  of  shadow.  The  plane  *of  rays, 
whose  horizontal  trace  is  hn^  parallel  to  the  horizontal 
projection  of  the  light,  determines  the  point  of  shadow 
(n,  w ; and  the  points  of  shadow  (o,  o'),  (^r,  (5,  /) 

(/,  f w')  and  (t;,  i;')  are  found  in  a similar  manner. 

The  vertical  plane  of  rays,  whose  horizontal  trace  is 
mu,  is  tangent  to  the  column  along  the  element  of  shade. 
The  shadow  ends  and  the  shade  begins  at  the  point  (u,u'). 

The  shade  on  the  abacus  is  found  by  drawing  a plane 
of  rays  which  shall  be  tangent  to  it — the  line  p drawn 
tangent  to  cmd  and  parallel  to  the  harizontal  projection 
of  the  rays  of  light,  is  the  horizontal  trace  of  the 
tangent  plane — the  element  of  contact  is  the  element 
of  shade. 

The  part  of  the  surface  above  the  line  of  shadow,  as 
well  as  that  portion  of  it  which  is  in  the  shade,  is  shaded 
in  the  drawing. 

PROBLEM  VII. 

Having  given  a cylindrical  abacus  and  the  direction  of  the 
lights  it  is  required  to  find  the  shadow  of  the  abacus  on  a 
vertical  plane^  or  wall. 

32.  Let  anbp  (PI.  6 Fig.  3)  be  the  plan,  and  a'cd'U 
the  elevation  of  the  abacus,  the  abacus  touching  the 
vertical  plane  on  which  the  shadow  falls,  in  the  element 
{p,  no). 


r.. 


9-  -- 

“O 


w 


IT 


W: 

..J  . ' 

,3  - 


% ■ 


SHADES  AND  SHADOWS, 


35 


Suppose  t^vo  squares  to  be  constructed,  the  one  cir- 
cumscribing the  upper  circle  of  the  abacus,  the  other 
the  lower.  The  square  cdgl  is  the  projection  of  both 
the  squares  on  the  horizontal  plane. 

Let  two  planes  of  rays  be  drawn  tangent  to  the  abacus. 
The  lines  /f"  and  ii  \ drawn  parallel  to  the  horizontal 
projection  of  the  ray  of  light  and  tangent  to  the  circle 
nhpa^  are  their  traces;  and  (^,  (^,  yx)  are  the 

elements  of  contact,  and  consequently  the  elements  of 
shade  (27). 

Now  the  lines  of  the  abacus  which  cast  lines  of 
shadow  on  the  vertical  plane,  are,  1st,  the  upper  semi- 
circle {thi^  b'ty') ; 2dly,  the  two  elements  of  shade  (/,  t's) 
(^,  yx^  ; and  3dly,  the  lower  semicircle  (iant^  cxs). 

Through  (z,  n ),  the  centre  of  the  upper  circle  of  the 
abacus,  let  a ray  of  light  be  drawn, — ^the  point  z\  in 
which  it  pierces  the  vertical  plane,  is  the  centre  of  the 
ellipse  of  shadow  cast  by  the  upper  circle.  The  ray 
through  («,  w ) pierces  the  vertical  at  jt?,  hence  nzp  is 
the  shadow  cast  by  the  diameter  {np^  n').  Drawing 
rays  through  (6,  6')  and  (a,  d')  determines  w and  i;,  the 
extremities  of  the  shadow  cast  by  the  diameter  (a6. 
Hence  np  and  vw  are  conjugate  diameters  of 
the  ellipse  of  shadow  (28). 

Drawing  rays  of  light  through  the  points  (J,  6')  and 
(c,  a),  determines  bf  the  shadow  cast  by  the  tangent 
(o(g,  by,  also  ef  the  shadow  cast  by  the  tangent  {cd, 
dby  and  de  the  shadow  cast  by  the  tangent  {cl,  a'). 
But  since  these  lines  are  all  tangent  to  the  upper  circle 
of  the  abacus,  their  shadows  will  be  tangent  to  the 
ellipse  of  shadow  (16) ; and  the  same  maybe  shown  of 
the  lower  circle  of  the  abacus  and  its  tangents. 

The  ellipse  of  shadow  cast  by  the  lower  circle  of  the 
abacus  is  easily  found.  The  point  z"  is  its  centre,  oq. 


36 


TREATISE  ON 


xy  are  its  conjugate  diameters,  and  the  lines  cTc, 
cg^  and  gh  are  tangent  to  it.  The  elements  of  shade 
cast  the  shadows  fu  and  ^7.  Hence,  the  lines  of  shadow 
on  the  vertical  plane,  are — the  semi-ellipse  Ixqu^  cast 
by  the  lower  semicircle  {iant^  cxs) ; the  right  line  ui\ 
cast  by  the  element  (/,  7^);  the  semi-ellipse  fwni\ 
cast  by  the  upper  semicircle  {tbi^  b't'y),  and  the  right  line 
i'l  cast  by  the  element  of  shade  (^,  yx^, 

PROBLEM  VIII. 

Having  given  the  frustum  of  an  inverted  cone^  it  is  re- 
quired to  find  the  shadow  cast  by  the  upper  circle  on  the  inner 
surface^  and  the  shadow  cast  by  the  frustum  on  the  plane  of  the 
lower  base, 

33.  Let  the  circle  CLHD  (PI.  4)  be  the  horizontal  pro- 
jection of  the  upper  circle  of  the  frustum,  and  G'H  its 
vertical  projection.  Let  the  circle  described  with  the 
centre  A and  radius  AK  be  the  horizontal,  and  TK'  the 
vertical  projection  of  the  lower  base.  The  axis  of  the 
cone  being  supposed  perpendicular  to  the  horizontal 
plane,  is  projected  on  the  horizontal  plane  at  A,  and  on 
the  vertical  plane  in  the  line  A' A^  perpendicular  to  the 
ground  line,  and  Gl'K'H  is  the  vertical  projection  of  the 
frustum.  Producing  the  line  G'l'  till  it  intersects  A" A', 
gives  A,  the  vertical  projection  of  the  vertex  of  the 
cone  of  which  the  frustum  is  a part. 

To  find  the  shadow  on  the  inner  surface  . 

If  two  tangent  planes  of  rays  be  drawn  to  the  cone, 
the  points  in  which  the  elements  of  contact  meet  the 
upper  circle,  are  the  points  where  the  shadow  on  the 
inner  surface  begins. 

Through  (A,  A),  the  vertex  of  the  cone,  let  a ray  of 


SHADES  AND  SHADOWS. 


37 


light  be  drawn.  Such  ray  pierces  the  plane  of  the  upper 
base  of  the  frustum  in  the  point  (B,  B').  Through  the 
point  (B,  B'),  suppose  two  lines  to  be  drawn  tangent  to 
the  upper  circle  of  the  frustum.  These  tangents  are  the 
traces,  on  the  plane  of  the  upper  base,  of  two  planes  of 
rays  drawn  tangent  to  the  cone.  The  lines  BC,  BD, 
drawn  tangent  to  the  circle  GDHL,  are  the  horizontal 
projections  of  these  traces;  and  (D,  D')  (C,  C')  are  the 
points  at  which  the  shadow  on  the  inner  surface 
begins. 

Let  the  surface  of  the  cone  be  now  intersected  by 
planes  of  rays  passing  through  the  vertex.  Each  se- 
cant plane"  so  drawn  will  intersect  the  surface  in  two 
elements ; the  element  towards  the  source  of  light 
will  cast  the  shadow,  and  the  other  will  receive  it. 
Each  plane  will  also  contain,  the  ray  of  light  passing 
through  the  vertex  of  the  cone,  and  consequently  every 
trace  on  the  plane  of  the  upper  base  will  pass  through 
the  point  (B,  B'). 

Draw  any  line,  as  BaL,  to  represent  the  trace  of  such 
a plane.  The  horizontal  projections  of  the  elements  in 
which  it  intersects  the  surface  of  the  cone,  are  AL 
and  Acf.  Projecting  L and  a into  the  upper  base  at  U 
and  « , and  joining  these  points  with  A,  the  vertical 
projection  of  the  vertex  of  the  cone,  gives  the  vertical 
projections  of  these  elements. 

Through  (a,  a ) the  upper  extremity  of  the  element 
towards  the  source  of  light,  conceive  a ray  of  light  to 
be  drawn ; the  point  (^,  6')  in  which  it  meets  the  ele- 
ment (AL,  AL')  is  a point  of  the  curve  of  shadow. 

If  we  suppose  the  surface  of  the  frustum  to  be  pro- 
duced below  the  plane  E'F',  the  shadow  that  w^ould  fall 
on  the  part  of  the  surface  below  this  plane  is  easily 
found.  The  lowest  point  is  determined  by  passing  a 


38 


TREATISE  ON 


plane  of  rays  through  the  axis  of  the  cone.  The  trace 
of  this  plane,  on  the  upper  base  of  the  frustum,  is  hori- 
zontally projected  in  the  line  BAg?.  The  line  Kd!  is  the 
vertical  projection  of  the  element  which  receives  the 
shadow,  and  p)  is  the  point  of  shadow  cast  by  (^,  K), 

By  passing  a plane  of  rays  through  the  element  (AH, 
A'll)  we  shall  find  the  point  at  which  the  vertical  projec- 
tion of  the  shadow  is  tangent  to  the  element  A'H.  The 
horizontal  projection  of  the  trace  of  this  plane,  on  the 
upper  base  of  the  cone,  is  B^H.  Projecting  k into  the 
vertical  plane  at  k\  and  drawing  k'l  parallel  to  the  ver- 
^ tical  projection  of  the  light,  gives  I for  the  point  of  tan- 
gency.  The  horizontal  projection  of  the  point  is  found 
by  projecting  'i  into  the  horizontal  projection  of  the  ele- 
ment at  I Having  found  as  many  points  of  the  curve 
as  are  necessary  to  describe  it  accurately,  let  its  pro- 
jections be  made  as  in  the  figure. 

It  is  to  be  observed,  that  the  shadow  on  the  inner 
surface  of  the  frustum  intersects  the  plane  of  the  lower 
base  at  the  points  (c,  c ),  (/,  f) ; and  that  the  curve  of 
shadow  below  this  plane  is  found  on  the  supposition  that 
the  plane  does  not  intercept  the  light,  and  that  the  frusr 
turn  is  produced  below  it. 

If  the  plane  of  the  lower  base  be  supposed  to  inter- 
cept the  light,  there  will  be  no  shadow  on  the  surface 
of  the  cone  below  it ; for  the  plane  itself  will  receive 
the  shadow  cast  by  the  upper  circle. 

Through  the  centre  (A,  A'")  of  the  upper  circle,  con- 
ceive a ray  of  light  to  be  drawn.  It  pierces  the  plane 
of  the  lower  base  at  (F,  F').  A circle  described  with  F 
as  a centre,  and  radius  equal  to  the  radius  of  the  upper 
base,  is  the  horizontal  projection  of  the  shadow  cast  by 
the  upper  circle  on  the  plane  E'F'.  The  arc  cqf  falls 
y^ithin  the  circle  of  the  lower  base,  and  therefore  is  the 


SHADES  AND  SHADOWS. 


39 


line  of  shadow,  within  the  surface,  on  the  plane  of  that 
base.  The  circle  described  with  the  centre  F will  pass 
through  c and  /,  the  horizontal  projections  of  the  points 
in  which  the  curve  of  shadow  intersects  the  plane  ET'. 

The  shadow  cast  by  the  frustum  on  the  plane  of  the 
lower  base  is  limited  by  the  traces  of  the  two  tangent 
planes  of  rays  that  determine  the  elements  of  shade. 
These  planes  contain  the  ray  of  light  drawn  through 
the  vertex ; hence  their  traces  pass  through  the  point 
(E,  E')  in  which  this  ray  pierces  the  plane  E'F' ; and  Ew, 
Em  drawn  through  E and  tangent  to  the  circle  ^Kc  are 
their  horizontal  projections. 

The  shadow^s  cast  by  the  elements  of  shade  on  the 
plane  ET'  begin  at  the  points  in  which  they  pierce  it, 
and  terminate  at  the  points  m and  w,  where  the  shadows 
are  tangent  to  the  shadow  cast  by  the  upper  circle.  The 
points  m and  n may  be  determined  by  drawing  horizon-  - 
tal  projections  of  rays  through  the  points  C and  D,  and 
noting  their  intersections  with  the  shadow  cast  by  the 
upper  circle. 

That  part  of  the  shadow  on  the  plane  ET'  which  is 
under  the  surface  of  the  cone  is  not  seen  in  horizontal 
projection. 

For  the  purpose  of  showing  the  appearance  of  the 
shadow  on  the  inner  surface,  we  will  suppose  that  part 
of  the  frustum  which  is  in  front  of  the  vertical  plane  GAH 
to  be  removed,  and  then  represent  in  vertical  projection 
the  remaining  semi-frustum,  as  it  appears  to  the  eye. 

That  part  of  the  surface  which  lies  between  the  ele- 
ment (AG,  AG')  and  the  curve  of  shadow,  being  in  the 
shadow,  is  darkened  in  vertical  projection. 


40 


TREATISE  ON 


PROBLEM  IX. 

Having  given  a sphere  in  space^  and  the  direction  of  the 
lights  it  is  required  to  find  the  curve  of  shade^  and  the  shadows 
cast  on  the  planes  of  projection, 

34.  Let  the  centre  of  the  sphere  be  taken  at  equal  dis- 
tances from  the  planes  of  projection ; let  A (PI.  5)  be  its 
horizontal' and  A its  vertical  projection;  and  suppose 
the  projections  of  the  light  to  make  equal  angles  with 
the  ground  line. 

Suppose  the  sphere  to  be  circumscribed  by  a tangent 
cylinder  of  rays.  The  axis  of  the  cylinder  is  the  ray  of 
light  passing  through  the  centre  of  the  sphere ; the 
curve  of  contact  is  a great  circle  whose  plane  is  perpen- 
dicular to  the  axis  of  the  cylinder,  that  is,  to  the  direc- 
tion of  the  light  in  space.  This  circle  of  contact  is  the’ 
curve  of  shade. 

Since  the  axis  of  the  tangent  cylinder  of  rays  is 
oblique  to  both  planes  of  projection,  the  plane  of  the 
circle  of  shade  which  is  perpendicular  to  it,  is  also 
oblique  to  both  the  planes  of  projection,  and  conse- 
quently its  projections  on  these  planes  will  be  ellipses, 
— (Des.  Geom.  180.) 

To  find  the  projection  of  the  circle  of  shade  on  the 
horizontal  plane. 

The  horizontal  projection  of  that  diameter  of  the  cir 
cle  of  shade  which  in  space  is  parallel  to  the  horizontal 
plane,  is  the  transverse  axis  of  the  ellipse  into  which  the 
circle  of  shade  is  projected.  The  projection  of  that 
diameter  which  makes  the  greatest  angle  with  the  hori- 
zontal plane,  or  which  is  perpendicular  to  the  parallel 
diameter,  is  the  conjugate  axis.— (Des.  Geom.  180.) 


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SHADES  AND  SHADOWS. 


41 


The  ray  of  light  through  the  centre  of  the  sphere 
being  perpendicular  to  the  plane  of  shade,  is  also  per- 
pendicular to  that  diameter  of  the  circle  of  shade  which 
is  parallel  to  the  horizontal  plane,  and  therefore  the  pro- 
jections of  this  diameter  and  the  ray  of  light,  on  the  hori- 
zontal plane,  are  at  right  angles  to  each  other.  Hence 
the  diameter  CAD,  drawn  through  the  centre  A,  and  per- 
pendicular to  AB,  the  horizontal  projection  of  the  ray 
through  the  centre,  is  the  transverse  axis  of  the  ellipse 
into  which  the  circle  of  shade  is  projected. 

Through  the  centre  of  the  sphere  suppose  a plane 
of  rays  to  be  drawn  perpendicular  to  the  horizontal 
plane.  Its  horizontal  trace  is  EAB.  This  plane  inter- 
sects the  surface  of  the  sphere  in  a great  circle,  the 
cylinder  of  rays  in  two  elements  tangent  to  this  circle, 
and  the  plane  of  the  circle  of  shade  in  a diameter 
passing  through  the  points  of  contact.  The  projection 
of  this  diameter  is  the  conjugate  axis  of  the  ellipse. 

To  find  the  conjugate  axis  : 

Let  this  plane  be  revolved  about  a vertical  axis  pass- 
ing through  the  point  A,  till  it  becomes  parallel  to  the 
vertical  plane  of  projection.  In  the  revolution,  the  point 
B describes  in  the  horizontal  plane, the  arc  BB',  and  the 
point  (A,  A)  being  in  the  axis,  remains  fixed.  Project- 
ing B'  into  the  ground  line,  and  drawing  AB",  we  have 
the  vertical  projection  of  the  revolved  ray.  But  the 
great  circle  cut  out  of  the  sphere  is,  when  revolved  par- 
allel to  the  vertical  plane,  projected  into  the  circle  rep- 
resenting the  vertical  projection  of  the  sphere,  and  the 
elements  cut  from  the  cylinder  of  rays  are  projected 
into  lines  parallel  to  AB". 

Drawing  therefore  the  tangents  F'H'  and  G'b'  parallel 
to  AB",  we  determine,  in  the  revolved  position  of  the 
plane  of  rays,  the  extremities  of  that  diameter  of  the 
circle  of  shade  whose  projection  is  the  conjugate  axis. 


42 


TREATISE  ON 


When  the  plane  of  rays  is  parallel  to  the  vertical  plane, 
the  point  F'  is  horizontally  projected  at  F,  and  the  point 
G'  at  G.  In  the  counter-revolution  the  points  (F,  F') 
(G,  G')  describe  the  horizontal  arcs  (F/,  F/'),  (G^, 
G'^'),  and  gf  becomes  the  horizontal  projection  of  the 
diameter,  or  conjugate  axis  of  the  ellipse.  Having 
found  the  two  axes  of  the  ellipse,  let  it  be  described. 

It  is  plain  that  (/,/')  is  the  highest  point  of  the  curve 
of  shade,  and  (g’, g)  the  lowest ; therefore  the  tangents 
to  the  curve  of  shade  at  these  points  are  horizontal, 
and  their  vertical  projections  parallel  to  the  ground  line. 

Since  the  points  which  are  horizontally  projected  at 
C and  D are  contained  in  the  horizontal  plane  through 
the  centre  of  the  sphere,  they  will  be  vertically  projected 
in  its  trace  N'B'",  at  the  points  C'  and  D'. 

The  points  which  are  horizontally  projected  at  the 
intersections  of  the  ellipse  Cg-iy,  with  the  line  NB',  are 
vertically  projected  in  the  circumference  of  the  circle 
representing  the  vertical  projection  of  the  sphere.  We 
have  thus  found  six  points  of  the  ellipse  into  which  the 
circle  of  shade  is  projected  on  the  vertical  plane.  And 
since  the  tangents  passing  through  the  highest  and  the 
lowest  points  f and  g are  parallel  to  the  ground  line,  it 
follows  that  the  diameter  passing  through  these  points  is 
conjugate  with  the  diameter  D'C';  therefore  let  the 
curve  be  described  (29). 

The  axes  of  the  ellipse  into  which  the  circle  of  shade 
is  projected  on  the  vertical  plane  can,  however,  be  found 
by  a construction  similar  to  that  used  for  the  horizontal ' 
projection. 

Through  A'  draw  a diameter  perpendicular' to  AB, 
the  vertical  projection  of  the  ray.  This  will  be  the 
transverse  axis  of  the  ellipse.  Through  the  centre  of 
the  sphere  conceive  a plane  of  rays  to  be  drawn  per-^ 


SHADES  AND  SHADOWS. 


43 


pendicular  to  the  vertical  plane — A'B  is  its  vertical  trace. 
Let  this  plane  be  revolved  about  an  axis  passing  through 
the  centre  of  the  sphere  and  perpendicular  to  the  ver- 
tical plane,  until  it  becomes  parallel  to  the  horizontal 
plane.  The  point  B describes  in  the  vertical  plane  the 
arc  BB'";  and  since  (A,  A)  remains  fixed,  AB"  is  the 
horizontal  projection  of  the  revolved  ray  passing  through 
the  centre  of  the  sphere.  The  lines  t-H,  hb'  drawn 
parallel  to  the  revolved  ray  and  tangent  to  the  circle 
NDC,  determine  (c,  c),  (^,  /?-'),  the  extremities  of  the 
diameter,  in  its  revolved  position,  whose  vertical  projec- 
tion is  the  conjugate  axis  of  the  ellipse.  In  the  counter 
revolution,  the  points  (c,  c ) and  (A,  A')  describe  the  arcs 
(c  J,  cd!)  and  {h  Ulc) : and  k'd'  is  the  conjugate  axis  of 
the  ellipse  into  which  the  circle  of  shade  is  projected  on 
the  vertical  plane.  In  horizontal  projection,  we  see,  only 
that  part  of  the  shade  which  lies  above  the  horizontal 
plane  N'B'";  the  part  of  the  curve  of  shade  lying  below 
this  plane  is  dotted.  In  vertical  projection,  we  see  that 
part  of  the  shade  which  lies  in  front  of  the  plane  NB'. 

To  find  the  shadow  on  the  horizontal  plane : 

' The  curve  of  shadow  on  the  horizontal  plane  is  the 
intersection  of  the  horizontal  plane  with  the  surface  of 
' the  cylinder  of  rays  tangent  to  the  sphere.  This  curve 
will  be  an  ellipse,  unless  the  circle  of  shade  be  parallel 
to  the  horizontal  plane,  or  the  section  a sub-contrary 
one.  When  the  plane  of  rays  AB  was  revolved  par- 
allel to  the  vertical  plane,  we  drew  F'H'  and  G'b'  par- 
allel to  the  revolved  ray  A'B".  .The  tangent  (G'^',  Gb) 
pierces  the  horizontal  plane  at  b.  In  the  counter 
revolution,  the  point  b describes  in  the  horizontal  plane 
the  arc  b b'" ; then  b"'B  is  the  shadow  cast  by  the  radius 
of  the  circle  of  shade  which  passes  through  the  lowest 
point— hence  it  is  the  semi-transverse  axis  of  the  ellipse 


44 


TREATISE  ON 


of  shadow.  The  conjugate  axis  of  this  ellipse  is  the 
shadow  cast  by  the  horizontal  diameter  of  the  circle  of 
shade.  It  passes  through  B,  is  perpendicular  to  and 

equal  to  CD.  The  shadow  cast  on  the  vertical  plane  is 
found  in  a manner  entirely  similar. 

OF  BRILLIANT  POINTS. 

35.  When  a ray  of  light  falls  upon  a surface  which 
turns  it  from  its  course  and  gives  it  another  direction, 
the  ray  is  said  to  be  reflected.  The  ray,  as  it  falls  upon 
the  surface,  is  called  the  incident  ray,  and  after  it  leaves 
the  surface,  the  reflected  ray.  The  point  at  which  the 
reflection  takes  place  is  called  the  point  of  incidence. 
It  is  ascertained  by  experiment, 

1°.  That  the  plane  of  the  incident  and  reflected  rays  is 
always  normal  to  the  surface  at  the  point  of  incidence. 

2°.  That  at  the  point  of  incidence,  the  incident  and 
reflected  rays  make  equal  angles  with  the  tangent  plane 
or  normal  line  to  the  surface. 

If  therefore,  we  suppose  a single  luminous  point,  and 
the  light  emanating  from  it  to  fall  upon  any  surface  and 
to  be  reflected  to  the  eye,  the  point  at  which  the 
reflection  takes  place  is  called  the  brilliant  point. 
The  brilliant  point  of  a surface  is,  then,  the  point  at 
which  a ray  of  light  and  a line  drawn  to  the  eye  make 
equal  angles  with  the  tangent  plane  or  normal  line — the 
plane  of  the  two  lines  being  normal  to  the  surface. 

36.  The  rays  of  light  being  parallel,  and  the  place  of 
the  eye  at  an  infinite  distance,  the  brilliant  point  of  any 
surface  is  thus  found: 

Through  any  point  in  space  draw  a ray  of  light  and 
a line  to  the  eye,  and  bisect  the  angle  included  between 
them.  Then  draw  a tangent  plane  to  the  surface  and  per- 
pendicular to  the  bisecting  line — the  point  of  contact  is 


SHADES  AND  SHADOWS. 


45 


the  brilliant  point.  F or,  let  AC  (Fig.  n)  be  the  lino  drawn 
to  the  eye,  AB  the  ray  of  light,  AD  the  bisecting  line, 
and  P the  point  of  contact  of  a plane  passed  perpendi- 
cular to  AD,  and  tangent  to  the  surface.  Through  P 
let  there  be  drawn  a ray  of  light  PG, which  will  be  paral- 
lel to  AB,  the  line  PE  to  the  eye, which  will  be  parallel  to 
AC,  and  PF  the  normal  line  to  the  surface  at  the  point  P. 

Since  the  normal  PF  is  perpendicular  to  the  tangent 
plane,  it  is  parallel  to  the  bisecting  line  AD,  for  AD  is 
perpendicular  to  the  tangent  plane  by  construction.  But 
the  line  AD  is  in  the  plane  of  the  lines  BA,  AC,  and 
bisects  the  angle  BAC  : therefore  the  parallel  PF  lies  in 
the  plane  of  the  lines  GP,  PE,  and  bisects  the  angle 
GPE.  Hence  P is  the  brilliant  point. 

36.  To  apply  these  principles  in  finding  the  brilliant 
point  on  the  surface  of  a sphere. 

Suppose  the  eye  to  be  in  a line  perpendicular  to  the 
vertical  plane,  and  at  an  infinite  distance  from  it. 

Through  (A,  A),  the  centre  of  the  sphere,  suppose  a 
ray  of  light  to  be  drawn,  and  also  a line  to  the  eye.  F or 
the  purpose  of  bisecting  the  angle  included  between  these 
lines,  let  their  plane  be  revolved  about  (AP,  A)  the  line 
drawn  to  the  eye,  until  it  becomes  parallel  to  the  horizon- 
tal plane.  Any  point  of  the  ray,  as  (E,  E'),  will  describe 
an  arc  (Ee,  E'e  ) ; and  Ae  and  AP  are  the  horizontal  pro- 
jections of  the  lines  when  revolved  parallel  to  the  horizon- 
tal plane.  Bisect  then,  the  angle  P Ac  by  the  line  Kq : and 
from  any  point  of  the  axis,  as  P,  draw  the  line  Pqe.  After 
the  counter  revolution,  the  point  e is  horizontally  pro- 
jected at  E ; and  since  P remains  fixed,  Vqe  is  horizon- 
tally projected  in  PE ; the  point  q of  the  bisecting  line, 
is  projected  at  q^  and  Kq[  is  the  horizontal  projection  of 
the  bisecting  line.  Its  vertical  projection  is  A'E'. 

The  plane  drawn  perpendicular  to  this  bisecting  line, 


46 


TREATISE  ON 


and  tangent  to  the  sphere,  touches  the  surface  at  the 
point  where  the  bisecting  line  pierces  it- — that  is,  at 
(a',  a").  Consequently,  (a , a")  is  the  brilliant  point.  A 
similar  construction  would  determine  the  brilliant  point, 
if  the  eye  were  taken  in  a line  perpendicular  to  the  hori- 
zontal plane. 


PROBLEM  X. 

Having  given  an  ellipsoid  in  space,  and  the  direction  of  the 
light,  it  is  required  to  find  the  curve  of  shade,  and  the  shadow 
cast  on  the  horizontal  plane. 

37.  Let  the  horizontal  plane  be  taken  perpendicular  to 
the  axis  of  the  surface.  Let  A (PI.  6.  Fig.  1)  be  the 
horizontal  projection  of  the  axis,  and  A'B  its  vertical 
projection.  Let  us  suppose  the  ellipsoid  to  be  circum- 
scribed by  a tangent  cylinder  of  rays.  The  axis  of  the 
cylinder  is  a ray  ofjight  passing  through  the  centre 
of  the  ellipsoid  and  piercing  the  horizontal  plane  at  C. 
Through  this  axis  let  a plane  of  rays  be  passed  perpen- 
dicular to  the  horizontal  plane — ACD  is  its  trace. 

Since  this  plane  is  a plane  of  rays,  and  divides  the  ellip- 
soid into  two  equal  and  symmetrical  parts,  the  parts  of 
the  curve  of  shade  lying  on  either  side  are  equal  and 
symmetrical.  Hence,  both  parts  are  projected  on  the 
meridian  plane  AC,  into  the  same  right  line.  But,  since 
the  contact  of  the  ellipsoid  and  surface  of  the  cylinder 
is  an  ellipse  whose  plane  passes  through  the  centre  of 
the  ellipsoid,  the  curve  of  shade  is  projected  into  a right 
line  passing  through  the  centre,  which  line  is  the  inter 
section  of  the  plane  of  the  curve  of  shade  with  the 
meridian  plane  ACD.  The  plane  x4CD  intersects  the 
surface  of  the  ellipsoid  in  a meridian  curve,  and  the  sur 


( 


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ii 


SHADES  AND  SHADOWS. 


47 


face  of  the  cylinder  of  rays  in  two  elements  that  are 
tangent  to  it.  The  points  of  tangency  are  the  highest 
and  lowest  points  of  the  curve  of  shade.  The  line  in 
which  the  plane  of  shade  intersects  the  meridian  plane 
AC,  also  passes  through  these  points. 

Let  the  ellipsoid  be  now  projected  on  a vertical  plane, 
parallel  to  the  meridian  plane  ACD.  Assuming  E'D' 
for  the  new  ground  line,  the  centre  of  the  ellipsoid  will 
be  projected  at  F,  in  the  perpendicular  AF  to  the  ground 
line  E'D',  and  at  a distance  from  it,  equal  to  the  distance 
of  the  centre  of  the  ellipsoid  from  the  horizontal  plane. 
The  ray  of  light  through  the  centre  will  be  vertically 
projected  in  FC'.  The  ellipse  described  about  the  cen- 
tre F,  and  equal  to  a meridian  curve  of  the  surface, 
represents  the  vertical  projection  of  the  ellipsoid. 

Having  described  this  ellipse,  draw  the  two  tangents 
d'T>'  and  eYI  parallel  to  FC',the  vertical  projection  of  the 
ray.  The  points  of  contact  d'\  e are  the  vertical  pro- 
jections of  the  highest  and  lowest  points  of  the  curve 
of  shade;  and  c?"Fe"  is  the  vertical  projection  of  the 
curve  of  shade,  since  the  plane  of  shade  is  perpendicu- 
lar to  the  new  vertical  plane.  The  horizontal  projec- 
tions of  the  highest  and  lowest  points  are  d and  e. 

Let  us  suppose  a system  of  horizontal  planes  to  inter- 
sect the  ellipsoid  between  the  highest  and  lowest  points 
of  the  curve  of  shade.  Each  of  such  secant  planes, 
being  perpendicular  to  the  axis,  will  intersect  the  sur- 
face of  the  ellipsoid  in  a horizontal  circle,  and  the  plane 
of  shade  in  a horizontal  line  perpendicular  to  the  meri- 
dian plane  ACD,  and  consequently,  to  the  new  vertical 
plane. 

Let/'?!  be  the  vertical  trace,  on  the  new  vertical  plane, 
of  one  of  the  secant  planes.  The  line/'?!  is  the  vertical 
projection  of  the  circle  in  which  the  plane  intersects  the 


48 


TREATISE  ON 


surface  of  the  ellipsoid,  and  K of  the  line  in  which  it 
intersects  the  plane  of  shade.  Projecting  the  circle 
and  line  on  the  horizontal  plane,  the  points  h and  in 
which  they  intersect,  are  two  points  in  the  horizontal 
projection  of  the  curve  of  shade.  In  a similar  mannei 
any  number  of  points  may  be  determined.  The  hori 
zontal  plane  through  the  centre  of  the  ellipsoid,  inter 
sects  the  plane  of  shade  in  a line  whose  horizontal  pro- 
jection is  and  this  line  is  the  transverse  axis-  of  the 
ellipse  into  which  the  curve  of  shade  is  projected ; and 
ed^  the  projection  of  the  diameter  joining  the  highest 
and  lowest  points,  is  the  conjugate  axis. 

. The  projection  of  the  curve  of  shade  on  the  primitive 
vertical  plane,  is  found  by  projecting  the  horizontal  cir- 
cles, and  drawing  perpendiculars  to  the  ground  line 
through  points  of  the  horizontal  projection  of  the  curve. 
Thus,  projecting  the  horizontal  circles  passing  through 
the  highest  and  lowest  points,  and  drawing  perpendicu- 
lars to  the  ground  line  from  d and  c,  determines  d'  and 
e,  the  vertical  projections  of  the  highest  and  lowest 
points.  The  vertical  projection  of  the  circle  fU'n^  is 
and  k'  are  the  vertical  projections  of  the  points  of 
shade  that  are  horizontally  projected  at  h and  k.  The 
diameters  d'e  and  t's  are  conjugate. 

The  part  of  the  surface  which  is  in  the  shade,  and  in 
front  of  the  meridian  plane  gAp,  is  shaded  in  vertical 
projection.  The  part  in  the  shade  and  above  the 
horizontal  plane  passing  through  the  centre  of  the  ellip- 
soid, is  shaded  in  horizontal  projection. 

To  find  the  shadow  on  the  horizontal  plane : 

If  we  suppose  the  lines  in  which  the  horizontal 
secant  planes  intersect  the  plane  of  shade,  to  cast 
shadows  on  the  horizontal  plane,  the  shadows  cast 
will  be  parallel  to  the  lines  themselves.  The  line  in 


SHADES  AND  SHADOWS. 


49 


which  the  horizontal  plane  f'n  intersects  the  plane  of 
shade,  casts  the  shadow and  the  points  W and  k"'m 
which  it  is  intersected  by  hh'"  and  k¥\  drawn  parallel  to 
the  horizontal  projection  of  the  light,  are  points  in  the 
shadow  on  the  horizontal  plane.  The  highest  and 
lowest  points  cast  their  shadows  at  D and  E. 

If  through  the  highest  and  lowest  points  of  the  curve 
of  shade,  two  planes  be  drawn  tangent  to  the  cylinder 
of  rays  circumscribing  the  ellipsoid,  their  traces  on  the 
horizontal  plane  will  be  tangent  to  the  ellipse  of  shadow ; 
but  since  the  planes  are  perpendicular  to  the  vertical  plane 
EAD,  their  traces  will  be  perpendicular  to  ED,  at  the 
points  E and  D.  Hence,  ED  is  an  axis  of  the  ellipse  of 
shadow.  The  other  axis  passes  through  the  middle 
point  C,  and  is  the  shadow  cast  by  the  diameter  {st^  sf). 

PROBLEM  XI. 

Having  given  the  plan  and  elevation  of  a niche^  and  the 
direction  of  the  light,,  it  is  required  to  fnd  the  shadow  which 
the  niche  casts  upon  itself 

38.  A niche  is  a recess  in  the  wall  of  a building.  It 
is  generally  composed  of  a semi-cylinder,  and  the  quad- 
rant of  a sphere  having  the  same  radius  as  the  base  of 
the  cylinder.  The  quadrant  of  the  sphere  rests  on  the 
upper  base  of  the  semi-cylinder,  forming  the  upper  part 
of  the  niche.  The  quadrant  and  semi-cylinder  are  tan- 
gent to  each  other  in  the  semicircle  which  separates  the 
cylindrical  from  the  spherical  surface. 

Let  AB  (PI.  7.  Fig.  1)  be  the  intersection  of  the  face 
of  a vertical  wall  with  a horizontal  plane,  and  the  semi- 
circle ACB  the  plan  of  the  niche.  The  rectangle  A"B' 
is  the  elevation ' of  the  cylindrical  part,  and  the  semi- 

D 


50 


TREATISE  ON 


circle  A'T'B"  of -the  spherical  part  of  the  niche — A'B" 
is  the  vertical  projection  of  the  semicircle  that  separates 
the  cylindrical  from  the  spherical  surface. 

The  lines  of  the  niche  which  cast  lines  of  shadow  on 
the  surface,  are, 

1”.  The  element  (A,  A A'). 

2®.  The  semicircle  AF'B". 

A part  of  the  shadow  cast  falls  on  the  base  of  the 
niche,  a part  on  the  cylindrical,  and  a part  on  the  spheri 
cal  surface. 

To  find  the  shadow  on  the  base  and  cylindrical  surface. 

Through  the  element  (A,  AA")  conceive  a plane  of 
rays  to  be  passed.  Its  horizontal  trace  AC  is  parallel  to 
the  horizontal  projection  of  the  rays  of  light,  and  the 
plane  intersects  the  cylindrical  surface  in  a second  ele- 
ment (C,  C'a),  opposite  the  source  of  light,  and  on  this 
element  the  shadow  falls.  The  line  A a,  drawn  through 
A parallel  to  the  vertical  projection  of  the  rays  of  light, 
limits  the  shadow  on  the  element : and  drawing  through 
C'  the  line  C'6,  parallel  to  the  vertical  projection  of  the 
ray  of  light,  determines  the  point  which  casts  a shadow 
at  C.  The  point  C is  common  both  to  the  cylindrical 
surface  and  to  the  base  of  the  niche.  Hence,  the  part 
Ab  of  the  element  (A,  A A)  casts  the  shadow  AC  on 
the  base  of  the  niche,  and  the  part  bA'\  the  shadow 
C'a  on  the  cylindrical  surface.  Above  the  point  a,  the 
shadow  on  the  cylindrical  surface  is  cast  by  the  semi- 
circle (AB,  AF'B"). 

Through  any  point  of  the  semicircle,  as  (F,  F'),  con- 
ceive a vertical  plane  of  rays  to  be  passed.  Its  hori- 
zontal trace  Ff  is  parallel  to  the  horizontal  projection 
of  the  rays  of  light,  and  the  plane  intersects  the  surface 
of  the  cylinder  in  the  element  (/,  ff')-  The  point  /", 
in  which  the  vertical  projection  of  the  ray  drawn  through 


SHADES  AND  SHADOWS. 


51 


F' meets  (//"),  is  the  vertical  projection  of  the  shadow 
cast  by  the  point  (F,F').  In  a similar  manner,  we  may 
determine  other  points  of  shadow  on  the  cylindrical 
surface. 

Above  the  line  A"B"  the  shadow  will  fall  on  the  spheri- 
cal surface.  Before  finding  this  shadow  we  will  demon- 
strate, that  when  a cylinder  intersects  a sphere,  the 
curve  in  which  it  enters  on  the  one  side,  is  equal  to  the 
curve  in  which  it  leaves  the  sphere  on  the  other. 

For,  the  parts  of  the  elements  of  the  cylinder  inter- 
cepted between  the  points  at  w^hich  they  enter,  and  the 
points  at  which  they  leave  the  sphere,  are  parallel  chords 
of  the  sphere.  Conceive  a plane  to  be  passed  through 
the  middle  point  of  one  of  these  chords  and  perpendicular 
to  it.  Such  plane  passes  through  the  centre  of  the 
sphere  and  bisects  all  the  other  parallel  chords.  Hence, 
the  curves  in  which  the  cylinder  enters  and  leaves  the 
sphere,  are  s)unmetrical  with  respect  to  this  plane,  and 
are  consequently,  equal.  Therefore,  if  one  of  them  is 
a circle,  the  other  will  be  an  equal  circle ; and  hence, 
when  the  surface  of  a cylinder  intersects  that  of  a sphere 
in  a great  circle,  it  will  at  the  same  time  intersect  it  in 
a second  great  circle,  and  the  planes  of  these  circles 
will  intersect  each  other  in  a diameter  of  the  sphere. 
The  two  elements  of  the  cylinder,  passing  through  the 
extremities  of  this  common  diameter,  are  perpendicular 
to  it,  since  they  are  tangent  to  the  sphere. 

Let  us  suppose  an  entire  hemisphere  to  be  described 
on  the  diameter  (AB,  A"B"),  having  the  plane  of  its 
great  circle  vertical.  Through  this  circle  suppose  a 
cylinder  of  rays  to  be  passed.  Half  the  surface  of  this 
cylinder  intersects  the  surface  of  the  hemisphere  in  a 
semicircle,  which  is  the  shadow  that  the  semicircle 
(AB,  A'F'B")  would  cast  on  the  hemisphere. 


52 


TREATISE  ON 


The  plane  of  the  circle  casting  the  shadow,  and  the 
plane  of  the  circle  of  shadow,  intersect  each  other  in  a 
line  passing  through  the  centre  of  the  hemisphere  and 
perpendicular  to  the  elements  of  the  cylinder  of  rays, 
that  is,  perpendicular  to  the  direction  of  the  light.  But, 
since  this  diameter  is  a line  of  the  vertical  plane  AB,  it 
is  parallel  to  the  vertical  plane  of  projection — hence,  its 
vertical  projection  is  at  right  angles  to  the  vertical  pro- 
jection of  the  rays  of  light  (Des.  Geom.  49).  There- 
fore, if  through  D',  we  draw  the  line  " perpendicular 
to  the  projection  of  the  ray,  we  determine  I/f,  the  ver- 
tical projection  of  the  diameter  in  which  the  plane  of  the 
circle  casting  the  shadow  intersects  the  plane  of  the 
circle  of  shadow.  The  semicircle  IK'nK  casts  the 
semicircle  of  shadow  on  the  hemisphere,  the  vertical 
projection  of  which  shadow  is  an  ellipse,  whose  trans- 
verse axis  is  \K  (Des.  Geom.  180). 

Through  (D,  D')  let  a plane  of  rays  be  passed  per- 
pendicular to  the  face  of  the  niche,  and  let  us  suppose 
for  a moment  the  vertical  plane  of  projection  to  be 
moved  forward  to  coincide  with  this  face.  The  point 
D'  will  then  be  directly  over  D,  and  nD'p  will  be  the  trace 
of  the  plane  of  rays. 

Let  us  now  assume  an  auxiliary  plane  of  projection, 
parallel  to  the  plane  whose  trace  is  n and  at  a given 
distance  from  it.  Draw  np  parallel  to  np^  to  represent 
the  trace  of  the  auxiliary  plane,  and  let  it  be  borne  in 
mind  that  this  trace,  as  well  as  wjo,  is  in  the  plane  of  the 
face  of  the  niche.  The  semicircle  in  which  the  plane 
of  rays  np  intersects  the  hemisphere,  is  projected  on 
the  auxiliary  plane  in  the  semicircle  nUlp,  The  pro- 
jections of  the  ray  of  light  passing  through  the  point 
(w,  w),  are  nlc  and  w'D'.  This  ray  is  projected  on  the 
auxiliary  plane  by  projecting  n into  the  trace  at  n\  and 


SHADES  AND  SHADOWS. 


53 


projecting  any  other  point  of  the  ray,  as  (/c,  D')  (Des. 
Geom.  14):  this  is  done  by  laying  off  Wk  equal  to  Y>k 
The  line  nkl  is  the  projection  of  the  ray  on  the  aux 
diary  plane,  and  the  point  /,  in  which  it  meets  the  semi 
circle  nlp\  is  the  projection  of  a point  of  the  curve  of 
shadow.  But  since  D'A',  a line  in  the  plane  of  the  cir- 
cle of  shadow,  is  perpendicular  to  the  plane  of  rays 
whose  trace  is  np^  and  consequently  to  the  auxiliary 
plane,  it  follows,  that  the  plane  of  shadow  is  per- 
pendicular to  these  planes ; hence,  the  projection  of  the 
shadow  on  either  of  them  is  a right  line  (Des.  Geom. 
82).  But  D"  is  the  projection  of  A',  a point  of  the  curve 
of  shadow,  and  a second  point  is  projected  at  hence^ 
D'7  is  the  projection  of  the  shadow  on  the  auxiliary 
plane.  This  line  is  also  the  projection  of  the  intersec- 
tion of  the  plane  of  rays,  whose  trace  is  np^  with  the 
plane  of  the  circle  of  shadow. 

In  space,  therefore,  the  line  whose  projection  on  the 
auxiliary  plane  is  D'7,  is  perpendicular  to  the  diameter 
\K : hence,  D7'  its  projection  on  the  vertical  plane,  is 
the  semi-conjugate  axis  of  the  ellipse  into  which  the 
circle  of  shadow  is  projected  (Des.  Geom.  180).  Let 
the  ellipse  be  described.  From  K to  the  point  q\  where 
the  ellipse  intersects  at  A"B",  the  shadow  falls  on  the  sur- 
face of  the  sphere.  . From  q the  shadow  falls  on  the 
cylindrical  surface.  The  arc  /i'E  casts  the  shadow  on 
the  spherical,  and  the  arc  A"E  the  shadow  on  the  cylin- 
drical surface. 

Had  the  quadrant  of  the  sphere  below  the  plane  A"B" 
been  permitted  to  intercept  the  light,  the  shadow  af"q\ 
instead  of  being  on  the  cylinder,  would  have  been  the 
shadov/  qVl  on  the  surface  of  the  sphere.  We  have  also 
dotted  the  line  of  shadow  aevq^  which  the  front  circle 


54 


TREATISE  ON 


of  the  hemisphere  would  cast  upon  the  vertical  cylinder, 
if  the  sphere  did  not  intercept  the  rays  of  light. 

The  part  of  the  surface  of  the  niche  lying  between 
the  lines  A'A"E/^'  and  the  line  of  shadow,  is  shaded  in 
vertical  projection. 

It  is  not  necessary  to  find  the  semi-conjugate  axis  of 
the  ellipse  into  which  the  circle  of  shadow  is  projected. 
We  may  find  points  of  the  curve,  and  describe  it  through 
them. 

Let  i's  be  the  vertical  trace  of  a plane  of  rays  per- 
pendicular to  the  face  of  the  niche.  This  plane  inter- 
sects the  hemisphere  in  a semicircle  which  is  projected 
on  the  auxiliary  plane  in  the  semicircle  described  with 
the  radius  D"r.  Draw I's'  parallel  to  n'L  The  point  s\ 
where  it  nieets  the  semi-circle  and  the  line  D'7,  is  the  pro- 
jection on  the  auxiliary  plane  of  a point  of  the  circle  of 
shadow.  This  point  is  projected  on  the  vertical  plane 
of  projection  at  s' ^ and  on  the  horizontal  plane  at  5^,  by 
making  ms  equal  to  ms".  In  a similar  manner  other 
points  of  the  curve  may  be  found. 

The  point  H is  projected  on  the  horizontal  plane  at  /?, 
and  the  point  q ^tq:  therefore,  hsq  is  the  horizontal  pro- 
jection of  the  shadow  which  falls  on  the  spherical  part 
of  the  niche.  The  shadow  on  the  cylindrical  part  is 
horizontally  projected  in  the  arc  Ceq. 

The  point  (9^,  9''),  at  which  the  shadow  passes  from 
the  cylindrical  to  the  spherical  part  of  the  niche,  can  be 
found  by  a direct  construction. 

The  point  in  space,  is  the  one  in  which  the  intersec- 
tion of  the  upper  base  of  the  cylinder  with  the  plane  of 
shadow,  meets  the  circle  of  the  upper  base.  But  hH  is 
the  trace  of  the  plane  of  shadow  on  the  face  of  the  niche, 
and  (5*,  5)  is  a point  of  this  plane.  Therefore,  we  have 
one  trace  of  an  oblique  plane,  and  a point  of  the  plane, 


SHADES  AND  SHADOWS. 


55 


to  find  its  other  trace  (Des.  Geom.  43).  Draw  su 
parallel  to  NU ; its  horizontal  projection  is  su,  and  (u,  u ) 
is  the  point  in  which  it  pierces  the  horizontal  plane  A"B": 
therefore,  (u,  u')  is  a point  in  the  trace  of  the  plane  of 
shadow.  But  the  trace  passes  through  (D,  D')  ; hence 
Dug  is  the  horizontal  projection  of  the  required  trace. 
But  q is  the  vertical  projection  of  g ; hence  (q,  q)  is 
the  point  at  which  the  shadow  passes  from  the  cylin- 
drical to  the  spherical  surface. ' 


PROBLEM  XII. 

To  find  the  curve  of  shade  on  the  surface  of  a torus. 

39.  Let  abcg  (PI.  7,  Fig.  2)  be  a rectangle,  having  the 
semicircles  aA'b  and  cB'g  described  on  its  vertical  sides. 

If  the  figure  aA'bcB'g  be  revolved  about  a vertical  axis 
(C,  C"0),  it  will  generate  a solid,  called  a torus.  This 
solid  is  used  in  some  of  the  orders  of  architecture  in 
forming  the  bases  and  capitals  of  the  columns. 

Before  finding  the  curve  of  shade,  we  will  demonstrate, 
that  if  a surface  of  revolution  be  intersected  by  a me- 
ridian plane,  and  a line  be  drawn  tangent  to  the  meridian 
curve  and  parallel  to  the  projection  of  the  rays  of  light 
on  the  meridian  plane,  the  point  of  contact  will  be  a 
point  of  the  curve  of  shade. 

F or,‘'every  plane  of  rays,  tangent  to  a surface,  touches 
it  in  a point  of  the  curve  of  shade.  But  since  the  surface 
is  one  of  revolution,  such  plane  is  perpendicular  to.,  the 
meridian  plane  passing  through  the  point  of  contact  (Des. 
Geom.  105).  Therefore,  its  trace  will  not  only  be  tangent 
to  the  meridian  curve,  but  also  parallel  to  the  projection 
of  the  rays  of  light,  since  the  rays  are  projected  by  per- 


56 


TREATISE  ON 


pendicular  planes.  Hence,  the  tangent  to  a meridian  curve^ 
drawn  parallel  to  the  projection  of  the  rays  of  light  on  its  plane^ 
determines  a point  of  shade. 

Let  the  circle  ATBH  represent  the  horizontal  pro- 
jection of  the  surface,  g'B'chKa  its  vertical  projection, 
and  (A,  A)  the  ray  of  light.  Through  any  point  of  the 
axis  of  the  surface,  as  (C,  C'),  suppose  a ray  of  light  to 
be  drawn,  CD,  C'D'  are  its  projections.  Through  this 
ray  let  a meridian  plane  be  passed,  and  then  revolved 
about  the  axis  of  the  surface  until  it  becomes  parallel  to 
the  vertical  plane  of  projection.  The  point  (C,  C')  being 
in  the  axis,  remains  fixed,  and  any  point  of  the  ray,  as 
(D,  D'),  describes  a horizontal  arc  (Dc?,  D'c?');  and  the 
vertical  projection  of  the  ray,  from  its  revolved  position, 
is  C'd',  The  section  of  the  surface,  when  revolved 
parallel  to  the  vertical  plane,  is  vertically  projected  in 
the  meridian  line  g'B'cbKa. 

If  now,  two  lines  be  drawn  tangent  to  the  semicircles 
and  parallel  to  C'c?',  the  revolved  ray,  they  will  determine 
the  points  (^,  Id')  and  (^,  i) : these  are  the  highest  and  low- 
est points  of  the  curve  of  shade,  in  their  revolved  position. 
In  the  counter  revolution  of  the  meridian  plane,  they 
describe  horizontal  arcs,  and  when  the  counter  revolu- 
tion is  completed,  are  horizontally  projected  at  h and  /, 
and  vertically  at  K and  V, 

The  lines  drawn  parallel  to  C'D',  the  vertical  projec- 
tion of  the  ray,  and  tangent  to  the  semicircles,  determine 
the  points  (^,  and(/?,p)  at  which  the  curve  of  shade 

intersects  the  meridian  plane  AB. 

The  points  (T,  t')  and  (H,  t")  of  the  curve  of  shade 
are  found  by  passing  two  planes  of  rays  tangent  to  the 
surface  and  perpendicular  to  the  horizontal  plane.  The 
planes  will  touch  the  surface  in  the  circumference  of  the 
circle  whose  vertical  projection  is  AB',  The  points  of 


SHADES  AND  SHADOWS.  57 

tangency  are  also  found  in  a meridian  plane  passed  per- 
pendicular to  the  plane  DCA. 

To  find  other  points  of  the  curve  of  shade  : 

Draw  any  meridian  plane,  as  wCm,  and  project  the  ray 
of  light  upon  it.  The  point  (C,  C')  is  its  own  projec- 
tion, and  (D,  D)  is  projected  at  (F,  F') ; therefore,  (CF, 
C'F')  is  the  projection  of  the  ray  on  the  meridian  plane 
nCm,  Let  the  meridian  plane  be  revolved  till  it  becomes 
parallel  to  the  vertical  plane  of  projection.  The  line  (CF, 
C'F')  will  then  be  vertically  projected  in  the  line  Cf,  Draw 
two  lines  parallel  to  C'f  and  tangent  to  the  semicircles ; 
their  points  of  contact  (r,  /)  and  (v^  v)  are  points  of  the 
curve  of  shade  in  their  revolved  position.  In  the  counter 
revolution  these  points  describe  horizontal  arcs,  and 
when  it  is  completed  are  horizontally  projected  at  n 
and  m,  and  vertically  at  7i  and  m\  Any  number  of  points 
of  the  curve  of  shade  may  be  found  in  a similar  manner. 
The  curve  HqlTphm  is  the  horizontal,  and  the 

vertical  projection  of  the  curve  of  shade. 


PROBLEM  XIIL 

Having  given  a surface  of  revolution^  convex  towards  the 
axis,  and  the  direction  of  the  light,  it  is  required  to  find  the 
shadow  which  the  upper  circle  casts  upon  the  surface,  and  also 
the  brilliant  point, 

40.  Let  DRN  (PI.  8,  Fig.  1)  be  the  horizontal,  and 
D'L  the  vertical  projection  of  a cylinder  or  pedestal,  on 
which  the  surface  rests. 

T ake  two-thirds  of  the  radius  AD,  and  lay  it  off  from 
A'  to  V,  on  the  vertical  line  A'B.  Through  v draw  the 
horizontal  line  vu,  which  meets  the  vertical  line  D'?^  at 
tt.  With  w as  a centre,  and  radius  wD',  describe  the 


58 


TREATISE  ON 


quadrant  D'H'X'.  Then  lay  off  vB  equal  to  one-third 
of  the  radius  A'D',  and  draw  B/'  parallel  to  AD',  and 
equal  to  two-thirds  of  it.  With  y as  a centre,  and  a radius 
equal  to  yX'  or  yl\  describe  the  quadrant  l^bV.  The 
two  quadrants  will  have  a common  tangent  line  at  the 
point  X',  which  will  be  vertical.  Now,  if  the  curve  D'H'XY' 
be  revolved  about  the  vertical  axis  (A,  A'B)  it  will  gene- 
rate a surface  of  revolution,  convex  towards  the  axis, 
and  concave  outwards.  The  circle  whose  radius  is  X'l; 
is  called  thSfeircle  of  the  gorge. 

Let  us  also  suppose  a cylinder,  having  the  radius  of  its 
base  equal  to  the  radius  of  the  upper  circle  of  the  sur- 
face, to  be  placed  on  the  surface. 

It  is  required  to  find  the  curve  of  shadow  which  the 
upper  circle  (ZaQS,  ZV)  casts  upon  the  surface,  under  the 
supposition  that  the  light  is  not  intercepted  by  the  sur- 
face. Admitting  this  supposition,  it  follows,  that  each 
point  of  the  upper  circle  casting  a shadow  will,  in  gene- 
ral, cast  two  points  of  shadow;  one  where  the  ray 
through  the  point  enters  the  surface,  and  the  other 
where  it  leaves  the  surface. 

Through  the  upper  circle  of  the  surface  suppose  a 
cylinder  of  rays  to  be  passed.  The  ray  (AC,  BC')  is  the 
axis  of  this  cylinder,  and  the  curve  in  which  it  intersects 
the  surface  of  revolution  is  the  curve  of  shadow  required. 

Through  the  axis  of  the  cylinder  of  rays  and  the  axis 
of  the  surface,  suppose  a meridian  plane  to  be  passed. 
Its  horizontal  trace  is  AC,  and  it  cuts  the  upper  circle 
of  the  surface  in  two  points,  one  of  which  (a,  a ) casts 
shadows  on  the  meridian  curves.  Let  this  plane  be 
revolved  parallel  to  the  vertical  plane  of  projection. 
After  it  is  so  revolved,  the  ray  of  light  will  be  vertically 
projected  in  Be , the  meridian  curves  in  the  curves  repre- 
senting the  vertical  projection  of  the  surface,  and  the 
point  («,  a ) at  Z'. 


IMat. 


-4 


V 


c» 


SHADES  AND  SHADOWS. 


59 


Through  1 draw  I'bd  parallel  to  the  revolved  ray  Be' ; 
the  points  b and  d in  which  it  intersects  the  meridian 
curves,  are  the  highest  and  lowest  points  of  the  curve 
of  shadow,  in  their  revolved  position.  In  the  counter 
revolution  the  points  b and  d describe  arcs  of  horizontal 
circles,  and  when  the  revolution  is  completed,  are  hori- 
zontally projected  at  h and  g*,  and  vertically  at  H and  g. 
These  points  may  also  be  found  by  drawing  through 
the  point  in  which  the  revolved  ray  I'bd  meets  the  axis, 
the  line  gvH  parallel  to  BC',  the  vertical  projection  of 
the  ray,  and  noting  its  intersection  with  the  horizontal 
lines  dg  and  bK ; and  then  projecting  the  points  g and 
K into  the  horizontal  plane  in  the  line  aAC. 

To  find  points  of  the  curve  between  the  highest  and 
lowest  points,  we  intersect  by  horizontal  planes.  Each 
horizontal  plane  will  intersect  the  surface  of  revolution, 
in  a circle,  and  the  surface  of  the  cylinder  of  rays  in  a 
circle  equal  to  the  upper  circle  of  the  surface  : the  points 
in  which  these  circles  intersect  are  points  of  the  curve 
of  shadow. 

Let  H'C'  be  the  trace  of  an  auxiliary  horizontal  plane. 
This  plane  cuts  the  axis  of  the  cylinder  of  rays  in  the 
point  (C,  C').  With  C as  a centre,  and  a radius  equal 
to  Bz',  describe  the  arc  pm ; this  will  be  the  horizontal 
projection  of  an  arc  of  the  circle  in  which  the  auxiliary 
horizontal  plane  intersects  the  surface  of  the  cylinder 
of  rays.  Projecting  H'  into  the  line  DA,  we  have  AH 
for  the  radius  of  the  circle  in  which  the  same  plane  inter- 
sects the  surface  of  revolution.  Describing  that  circle, 
and  noting  the  points  in  which  it  intersects  the  circle 
described  witl  the  centre  C,  we  find  m and  p^  the  hori- 
zontal projections  of  two  points  of  the  curve  of  shadow. 
These  points  are  vertically  projected  in  the  vertical 
trace  of  the  auxiliary  plane,  at  m and  p'.  Similar  con^ 


60 


TREATISE  ON 


structions  determine  the  points  (^,  i')^  (c,  c ),  (o,  o'),  and 
(w,  n).  The  points  (/,  i)  and  (c,  c')  are  in  the  circle  of 
the  gorge. 

That  part  of  the  curve  is  made  full,  in  vertical  pro- 
jection, which  is  in  front  of  the  meridian  plane  DA2',* 
and  the  part  of  the  surface  which  is  above  the  curve 
of  shadow  is  shaded.  The  elements  of  shade  on  the 
pedestal  are  (N,  N')  and  (R,  R') ; and  on  the  upper 
cylinder  (Q,  Q')  and  (S,  S').  • 

41.  To  find  the  brilliant  point  : 

Suppose  the  eye  to  be  situated  in  a perpendicular  to 
the  vertical  plane,  and  at  an  infinite  distance  from  it. 

Through  any  point  of  the  axis,  as  (A,  A'),  suppose  a 
ray  of  light  to  be  drawn,  and  also  a line  to  the  eye,  and 
the  angle  contained  between  them  to  be  bisected  as  in 
Art.  36.  The  bisecting  line  is  (AK,  A'K').  It  is  now 
required  to  pass  a plane  perpendicular  to  this  line  and 
tangent  to  the  surface ; the  point  of  contact  will  be  the 
brilliant  point. 

If  we  suppose  the  tangent  plane  to  be  drawn,  its  trace 
on  the  meridian  plane  passing  through  the  bisecting  line 
(AK,  A'K')  will  be  perpendicular  to  the  bisecting  line, 
and  tangent  to  the  meridian  curve.  Let  this  meridian 
plane  be  revolved  parallel  to  the  vertical  plane  of  pro- 
jection. The  bisecting  line  will  then  be  vertically  pro- 
jected in  the  line  Kk\  and  the  meridian  curve  in  the  curve 
D'H'/'.  Let  H'l  be  drawn  perpendicular  to  and 
tangent  to  the  curve  D'H'/' ; the  point  of  contact  (H,  H') 
is  the  revolved  position  of  the  brilliant  point..  Draw 
the  normal  H' V perpendicular  to  the  tangent,  or  parallel 
to  Nk'.  In  the  counter  revolution,  V being  in  the  axis 
remains  fixed,  and  H'  describes  the  arc  of  a horizontal 
circle.  After  the  counter  revolution,  the  bisecting  line 
is  vertically  projected  in  A'K',  and  the  normal  VH'  in 


SHADES  AND  SHADOWS. 


61 


V P',  parallel  to  A'K'.  Hence  F,  where  the  horizontal 
line  HP'  intersects  VF,  is  the  vertical  projection  of  the 
brilhant  point.  Its  horizontal  projection  is  at  P,  in  the 
horizontal  trace  of  the  meridian  plane  AK. 

The  construction  here  given,  is  general  for  all  sur- 
faces of  revolution.  If  the*  eye  is  supposed  to  be  in 
a line  perpendicular  to  the  horizontal  plane,  the  brilliant 
point  is  easily  found ; for  we  bisect  the  angle  as  before, 
and  draw  a tangent  plane  perpendicular  to  the  bisect- 
ing line. 

It  is  plain  that  a second  line  can  be  drawn  perpen- 
dicular to  k'A!,  produced  on  the  other  side  of  xA,  which 
shall  be  tangent  to  the  meridian  curve  zd.  A second 
tangent  plane  can  therefore  be  drawn  perpendicular  to 
the  bisecting  line,  and  the  point  of  contact  will  answer 
the  mathematical  conditions  of  a brilliant  point.  The 
point,  however,  will  be  on  that  part  of  the  surface  which 
is  not  seen  by  the  eye. 

PROBLEM  XIV. 

Having  given  a surface  of  revolution  and  the  direction  of 
the  lights  it  is  required  to  find  the  curve  of  shade. 

42.  Fig.  2,  PI.  8 represents  the  projections  of  a sur- 
face of  revolution  generated  as  in  the  last  problem. 

Every  ray  of  light  that  is  tangent  to  the  surface  of 
revolution  is  an  element  of  the  tangent  cylinder  of  rays 
which  determines  the  curve  of  shade ; therefore,  every 
point  at  which  a ray  of  light  is  tangent  to  the  surface 
is  a point  of  the  curve  of  shade. 

Through  the  axis  of  the  surface  let  a meridian  plane 
of  rays  be  drawn — PB  is  its  horizontal  trace.  Let  this 
plane  be  revolved  until  it  becomes  parallel  to  the  verti- 
cal plane  of  projection.  The  ray  of  light  through  (A,  A) 


62 


TREATISE  ON 


will  then  be  vertically  projected  in  the  line  Kb\  and  the 
meridian  curves,  in  the  curves  which  represent  the  ver- 
tical projection  of  the  surface.  Let  the  two  tangents 
dd  and  ge  be  drawn  parallel  to  the  revolved  ray  Nb' ; the 
points  of  contact  («,  a)  and  (c, e)  are  the  highest  and 
lowest  points  of  the  curve  of  shade  in  their  revolved 
position.  After  the  counter  revolution,  these  points  are 
horizontally  projected  at  c and  /,  and  vertically  at  c and 
The  lines  gf  and  dc  are  parallel  to  A'K  the  verti- 
cal projection  of  the  ray. 

To  find  other  points  of  the  curve  of  shade,  we  use 
auxiliary  tangent  surfaces. 

Draw  any  line,  as  ET,  between  the  highest  and  lowest 
points.  At  E'  or  /c , draw  a tangent,  as  E'C,  to  the  meridian 
curve.  Conceive  the  meridian  plane  and  the  right-angled 
triangle  E'CO  to  be  revolved  about  the  axis  of  the  sur- 
face. The  meridian  curve  generates  the  surface  of  revo- 
lution, and  the  line  E'C  the  surface  of  a right  cone  tan- 
gent to  it  in  a circle  whose  vertical  projection  is  E'A:',  and 
horizontal  projection  Em??.  If  now,  two  tangent  planes 
of  rays  be  passed  to  this  cone,  they  will  be  also  tangent 
to  the  surface  of  revolution  at  two  points  in  the  circum- 
ference of  the  circle  (Em??,  E'^')  : these  points  are  points 
of  the  curve  of  shade. 

Through  (A,  C),  the  vertex  of  the  cone,  let  a ray  of 
light  be  drawn.  This  ray  pierces  the  plane  of  the  cone’s 
base  at  (D,  D').  Through  (D,  D')  let  two  lines  be  drawn 
tangent  to  the  base  of  the  cone — these  tangents  are  the 
traces  of  the  two  planes  of  rays  that  are  tangent  to  the 
cone,  and  Dm  and  D??  are  their  horizontal  projections. 
Projecting  the  points  m and  n into  the  vertical  plane  at 
m and  ??',  we  determine  two  points  (m,  m')  and  (??,  w') 
of  the  curve  of  shade.  In  a similar  manner,  other 
points  of  the  curve  of  shade  may  be  found. 


SHADES  AND  SHADOWS. 


63 


If  the  circle  of  contact  (Emw,  E'^')  be  taken  nearer 
the  circle  of  the  gorge,  the  vertex  of  the  cone  will  be 
farther  from  the  horizontal  plane,  and  the  elements  will 
be  nearer  vertical.  And  when  the  circle  of  the  gorge 
is  assumed  for  the  circle  of  contact,  the  auxiliary  tan- 
gent surface  will  be  a vertical  cylinder,  having  a common 
axis  with  the  surface  of  revolution.  If  two  planes  of 
rays  be  drawn  tangent  to  this  cylinder,  the  elements  of 
contact  will  pierce  the  horizontal  plane  at  i and  /,  the 
opposite  extremities  of  a diameter  perpendicular  to  the 
horizontal  projection  of  the  ray  of  light.  This  cylinder 
determines  the  two  points  (/,  1')  and  (^,  z')  of  the  curve 
of  shade. 

There  is  yet  a third  method  of  finding  points  of  the 
curve  of  shade : it  is  by  means  of  auxiliary  tangent 
spheres. 

Assume  any  line,  as  jt?V,  for  the  vertical  projection  of 
the  circle  of  contact  of  a sphere  and  the  surface  of 
revolution.  At  p draw  a tangent  to  the  meridian  curve, 
and  pq  perpendicular  to  it.  The  point  q,  where  the 
perpendicular  meets  the  axis,  is  the  vertical  projection 
of  the  centre  of  the  sphere,  and  qp  is  its  radius.  Let 
us  now  suppose  this  sphere  to  be  circumscribed  by  a 
tangent  cylinder  of  rays.  This  cylinder  will  touch  the 
sphere  in  a great  circle,  the  plane  of  which  will  be  per- 
pendicular to  the  direction  of  the  light.  Therefore,  the 
trace  of  the  plane  of  the  circle  of  contact  of  the  sphere 
and  cylinder  of  rays,  on  the  meridian  plane  EA6,  or  on 
the  vertical  plane  of  projection,  will  be  perpendicular  to 
the  projection  of  the  ray  of  light  on  either  of  these 
planes  (Des.  Geom.  49).  Hence,  q s\  drawn  through 
q perpendicular  to  AB',  is  the  vertical  projection  of  the 
line  in  which  the  plane  of  the  circle  of  contact  of  the 
sphere  and  cylinder  intersects  the  meridian  plane  EA^. 


64 


TREATISE  ON 


But  the  plane  of  this  circle  of  contact  intersects  the 
plane  of  the  circle  of  contact  of  the  sphere  and  surface 
of  revolution  in  a horizontal  line  perpendicular  to  the 
direction  of  the  light.  This  horizontal  line  must  pierce 
the  meridian  plane  EA6  in  the  trace  q s\  and  also  m the 
trace  pr ; therefore  it  pierces  it  at  (5,  s).  Hence,  the 
line  tsv^  drawn  through  s perpendicular  to  the  horizontal 
projection  of  the  ray,  is  the  horizontal  projection  of  the 
intersection  sought.  Projecting  p into  the  horizontal 
plane  at  jo,  and  describing  a circle  with  ihe  radius  Ay?, 
we  have  the  horizontal  projection  of  the  circle  of  con- 
tact of  the  sphere  and  surface  of  revolution.  Projecting 
the  points  t and  in  which  the  line  tsv  intersects  this 
circle,  into  the  vertical  plane  at  t'  and  and  we  have  the 
points  (^,  /')  and  (t;,  t;')  which  are  common  to  the  surface 
of  revolution,  the  tangent  sphere,  and  the  cylinder  of  rays. 

If  through  these  points  planes  be  drawn  tangent  to 
the  cylinder  of  rays,  they  will  be  planes  of  rays,  and 
tangent  both  to  the  sphere  and  surface  of  revolution. 
Hence,  the  points  (/,  ^')  and  (v,  z/)  are  points  of  the  curve 
of  shade. 

If  we  suppose  the  space  within  the  surface  of  revo- 
lution to  be  unoccupied,  a part  of  the  curve  of  shade 
will  cast  a shadow  that  will  fall  on  the  convex  side  of  the 
surface. 

The  ray  of  light  which  determines  (c,  c ),  the  highest 
point  of  shade,  being  produced,  will  intersect  the  oppo- 
site meridian  curve : the  point  of  intersection  is  a point  of 
shadow.  There  are  points  of  the  curve  of  shade,  on  both 
sides  of  the  meridian  plane  PAD,  which  also  cast  sha- 
dows on  the  surface.  As  we  recede  from  the  meridian 
plane  PAD,  on  either  side,  the  part  of  the  ray  inter- 
cepted between  the  point  of  shade  and  the  correspond- 
ing point  of  shadow,  continually  diminishes,  and  finallv 


SHADES  AND  SHADOWS. 


65 


becomes  nothing,  or  the  point  of  shadow  unites  with  the 
point  of  shade  casting  it. 

At  these  two  points,  one  on  each  side  of  the  meridian 
plane  PAD,  the  ray  of  light  will  be  tangent  to  the  curve 
of  shade.  For,  when  the  point  of  shadow  unites  with 
the  point  of  shade,  they  become  consecutive  points  of 
the  curve  of  shade;  hence,  the  ray  passing  through 
them  is  tangent  to  it  (Des.  Geom.  65).  But  if  the 
rays  of  light  at  these  two  points  are  tangent  to  the  curve 
of  shade,  their  projections  will  be  tangent  to  the  projec- 
tions of  the  curve  (Des.  Geom.  90).  The  horizontal 
projection  of  the  curve  of  shade  being  constructed,  draw 
two  tangents  to  it  parallel  to  PAD,  the  horizontal  pro- 
jection of  the  ray  of  light.  The  points  of  contact  are  X 
and  I/,  and  Xcy  is  the  horizontal  projection  of  that  part 
of  the  curve  of  shade  which  casts  a shadow  on  the 
interior  of  the  surface.  Projecting  the  points  X and  y 
into  the  vertical  projection  of  the  curve,  at  X'  and  y\  and 
drawing  lines  parallel  to  the  vertical  projection  of  the 
rays,  the  lines  so  drawn  will  be  tangent  to  the  vertical 
projection  of  the  curve  of  shade. 

Descending  along  the  curve  of  shade,  from  the  points 
(X,  X),  (y,  y),  the  rays  of  light  touch  the  surface  on  the 
concave  side,  and  the  points  of  the  curve  of  shade  still 
cast  shadows  upon  the  surface.  When  we  reach  those 
points,  one  on  each  side  of  the  meridian  plane  PAD,  at 
which  the  point  of  the  curve  of  shade  unites  with  the 
point  of  the  curve  of  shadow,  the  rays  become  tangent 
to  the  curve.  Therefore,  drawing  two  other  tang^ts 
parallel  to  PD,  their  points  of  contact  w and  2 are  the 
horizontal  projections  of  two  points  at  which  the  rays 
of  light  are  tangent  to  the  curve  of  shade.  At  these 
points  the  curve  of  shade  returns  to  the  convex  side  of 
the  surface.  Projecting  these  points  into  the  vertical 

E 


66 


TREATISE  ON 


projection  of  the  curve  of  shade,  we  find  the  points  w 
and  z\  through  which,  if  the  projections  of  rays  be  drawn, 
they  will  be  tangent  to  the  vertical  projection  of  the 
curve  of  shade. 

That  part  of  the  curve  of  shade  whose  horizontal 
projection  is  yc^  lies  on  the  convex  side  of  the  surface ; 
the  part  ILvlmz  lies  on  the  concave  side  of  the  surface ; 
the  part  zfw  lies  on  the  convex  side,  and  the  part  wnity 
on  the  concave  side.  The  curve  is  symmetrical  with 
respect  to  the  meridian  plane  PAD. 

The  part  of  the  curve  of  shade  which  is  in  front  of 
the  meridian  plane  EA6  is  made  full  in  vertical  projec- 
tion, and  the  part  of  the  surface  lying  above  the  curve 
of  shade,  and  seen,  is  darkened. 

PROBLEM  XV. 

Having  given  the  position  of  a surface  of  revolution  and 
the  direction  of  the  lights  it  is  required  to  find  the  line  which 
separates  the  dark  from  the  illuminated  part  of  the  surface^ 
and  the  shadow  which  is  cast  on  the  horizontal  plane  of  pro- 
jection, 

43.  Let  (A,  A'B),  (PI.  9)  be  the  axis  of  the  surface, 
and  let  the  projections  of  the  surface  be  made  as  in  the 
figure. 

Find  the  shadow  which  the  upper  horizontal  circle 
CD  casts  upon  the  surface,  as  in  Prob.  13,  and  then 
find  the  curve  of  shade,  as  in  Prob.  14. 

The  highest  and  lowest  points  of  the  curves  of  shade 
and  shadow,  are  in  the  meridian  plane  of  rays  EAB'. 

By  considering  the  form  of  the  meridian  curves,  it  is 
plain,  that  the  highest  point  of  the  curve  of  shade  is 
above  the  highest  point  of  the  curve  of  shadow,  and  the 


4^  - 1 


> 


SHADES  AND  SHADOWS. 


67 


lowest  point  of  the  curve  of  shade  below  the  lowest  point 
of  the  curve  of  shadow.  These  curves  will  therefore 
intersect  each  other.  This  they  do  at  the  points  (a,  d) 
and  Q),  by  Above  these  points  the  curve  of  shadow 
being  below  the  curve  of  shade,  and  on  the  exterior  sur- 
face, separates  the  dark  from  the  illuminated  part  of  the 
surface. 

Below  these  points,  the  curve  of  shade  is  below  the 
curve  of  shadow,  and  separates  the  dark  from  the  illu- 
minated part  of  the  surface,  until  it  returns  to  the  con- 
vex side  of  the  surface  at  the  points  (c,  c)  and  (J,  dy 

It  has  already  been  observed,  in  Prob.  14,  that  the 
parts  of  the  curve  of  shade  (^fac,fdc)  and  (^gbd,  gb'dy 
which  are  on  the  concave  side  of  the  surface,  cast 
shadows  upon  it.  These  shadows  begin  at  the  points 
(c,  c')  and  ((/,  dy 

To  find  these  shadows,  we  will,  in  the  first  place,  find 
the  shadow  which  the  entire  curve  of  shade  would  cast 
on  the  horizontal  plane,  under  the  supposition  that  the 
surface  offers  no  obstruction  to  the  light.  This  is  done 
by  finding  where  rays  of  light,  drawn  through  the  several 
points  of  the  curve  of  shade,  pierce  the  horizontal  plane. 

The  upper  part  of  the  curve,  which  is  on  the  convex 
side  of  the  surface,  casts  the  shadow  f'd'g' — the  lower 
part  of  the  curve,  on  the  convex  side  of  the  surface, 
casts  the  shadow  cV'd” — the  parts  of  the  curve  on  the 
concave  side  of  the  surface,  cast  the  shadows  cff"  and 
d''xg\  Assume  now  any  horizontal  circle  below  the 
points  (c,  c ),  (c/,  c?') ; the  one,  for  example,  whose  ver- 
tical projection  is  and  whose  horizontal  projection 
is  the  circle  described  with  the  centre  A and  radius  Kh 
equal  to  mk\  The  centre  of  this  circle  casts  a shadow 
on  the  horizontal  plane  at  in — the  circumference  de- 
scribed with  rn  as  a centre,  and  radius  mp.  equal  to  nxk\ 


68 


TREATISE  OJS 


will  be  the  shadow  cast  on  the  horizontal  plane  by  the 
circumference  of  the  assumed  circle. 

If  through  the  points  p and  in  which  this  shadow 
intersects  the  shadow  cast  by  the  curve  of  shade,  rays 
of  light  be  drawn,  they  will  intersect,  in  space,  the  cir- 
cumference of  the  horizontal  circle  and  the  curve  of 
shade.  The  points  in  which  these  rays  intersect  the 
circumference  of  the  horizontal  circle,  are  points  of 
shadow  on  the  surface  which  are  cast  by  the  points  in 
which  the  rays  intersect  the  curve  of  shade. 

Therefore,  draw  the  horizontal  projections  of  rays 
through  the  points  p and  and  the  points  p and  q\  in 
-which  they  intersect  the  circumference  kpq\  described 
with  the  centre  A and  radius  A^,  are  the  horizontal 
projections  of  two  points  of  the  shadow  on  the  sur- 
face. These  points  are  vertically  projected  at  p and 
q.  By  similar  constructions,  we  may  find  any  num- 
ber of  points  in  the  shadow  which  the  curve  of  shade 
casts  on  the  surface. 

If  we  take  the  lower  horizontal  circle  FG,  we  shall 
find  the  points  (/,  /')  and  (^,  where  the  shadows  on 
the  surface  terminate.  The  curve  (cpt^  cpi^  and  (dqx^ 
d'qx')  may  now  be  described.  The  curve  (cpt^ 
intersects  the  meridian  plane  HA  at  the  point  (5, s). 

We  have  thus  found  the  lines  on  the  exterior  of  this 
surface,  which  separate  the  dark  from  the  illuminated 
part.  They  are  the  curve  of  shadow  until  it  intersects 
the  curve  of  shade,  then  the  curve  of  shade  until  it 
returns  to  the  interior  surface,  and  then'  the  curve  of 
shadow  cast  on  the  surface  by  the  curve  of  shade.  The 
light  does  not  fall  on  that  portion  of  the  surface  which 
is  above  and  within  the  parts  of  these  curves. 

Let  us  now  find  the  shadow  which  the  surface  casts 
on  the  horizontal  plane. 


SHADES  AND  SHADOWS. 


69 


The  elements  of  shade  (I,  F)  and  (K,  K')  cast  lines  of 
shadow  on  the  horizontal  plane,  which  are  found  by 
drawing  rays  of  light  through  their  upper  extremities. 
Then  find  the  shadow  cast  by  the  circumference  of  the 
circle  whose  vertical  projection  is  FG.  This  shadow 
intersects  the  shadow  cast  by  the  curve  of  shade  in  the 
points  t”  and  x'.  These  points  of  shadow  are  cast  by 
the  points  (/,  t’')  and  x'y 

Find  next,  the  shadow  cast  by  the  circumference  of 
the  upper  circle  of  the  surface.  The  centre  of  this  cir- 
cle casts  a shadow  at  B'.  With  B'  as  a centre,  and  a 
radius  equal  to  BD,  let  the  circumference  of  a circle  be 
described — this  circumference  is  the  shadow  sought. 

If  through  the  points  y and  2^,  in  which  this  shadow 
intersects  the  shadow  cast  by  the  curve  of  shade,  rays 
of  light  be  drawn,  they  will  intersect,  in  space,  both  the 
curve  of  shade  and  the  upper  circle  of  the  surface. 
These  points  of  shadow  are  therefore  cast  by  the 
points  (a,  a ) and  (5, 6')  in  which  the  curves  of  shade  and 
shadow  intersect  upon  the  surface.  The  shadows  fy 
and  x’z  are  cast  by  parts  of  the  curve  of  shade. 

The  elements  of  shade  (P,  P'),  and  (L,  L')  of  the 
upper  cylinder,  cast  shadows  which  are  tangent  to  the 
circumference  described  with  the  centre  B'.  The  cir- 
cumference of  the  upper  circle  of  the  cylinder  casts  a 
shadow  on  the  horizontal  plane,  which  is  also  tangent 
to  the  shadows  cast  by  the  elements  of  shade. 


OF  THE  SHADES  AND  SHADOWS  OP  THE  ROMAN  DORIC 
COLUMN. 

44.  This  column  is  composed  of  three  principal 
parts: — 1st.  The  base;  2d.  the  shaft;  and  3d.  the  capital. 
That  the  figure  may  not  be  too  complicated,  we  shall 


70 


TREATISE  ON 


first  find  the  shades  and  shadows  on  the  base  and  shaft 
of  the  column,  and  then  use  a separate  figure  to  deter- 
mine those  of  the  capital. 

45.  The  base  of  the  column  (PI.  10.  Fig.  1)  is  com- 

posed of  seven  parts.  1.  A rectangular  prism,  called 
a plinth,  whose  horizontal  sections  are  squares : 2d. 

A solid  of  revolution,  convex  outward,  called  the  lower 
torus : 3d.  A cylindrical  fillet;  4th.  A solid  of  revolution, 
concave  outward,  called  a scotia ; 5th.  A cylindrical 
fillet ; 6th.  The  upper  torus : 7th.  A cylindrical  fillet. 

The  shaft  of  the  column  rises  from  the  upper  fillet. 
For  a short  distance  it  is  concave  outward,  then  it  be- 
comes nearly  cylindrical,  and  continues  so  to  near  its 
upper  extremity,  where  it  again  becomes  concave  out- 
ward. The  drawing  is  made  on  the  supposition  that 
the  shaft  is  cylindrical  between  the  parts  of  it  which  are 
concave  outward. 

46.  The  capital  is  composed  of  two  distinct  parts- — 
the  one  a member  whose  horizontal  sections  are 
squares,  having  their  centres  in  the  axis  of  the  column, 
the  other  a solid  of  revolution.  The  projections  of 
the  capital  are  shown  in  Fig.  2.  The  part  of  the  capi- 
tal whose  vertical  projection  is  wwdss^  is  the  portion 
whose  horizontal  sections  are  squares — this  part  is 
called  the  abacus.  The  part  of  the  abacus  between 
rr  and  nn  is  called  the  cyma-reversa,  or  talon. 

The  part  of  the  capital  whose  vertical  projection  is 
IV ram  is  called  the  echinus.  The  part  whose  vertical 
projection  is  hh'U  is  called  the  upper  fillet.  The  part 
whose  vertical  projection  is  ii'Mk  is  called  the  cavetto. 
The  part  whose  vertical  projection  is  gg'Kh  is  called  the 
neck.  The  part  / f is  called  the  astragal,  or  colarino. 
The  part  whose  projection  is  deVee  is  called  the  lower 


SHADFS  AND  SHADOWS. 


71 


fillet.  The  entire  column  below  the  abacus  and  above 
the  plinth  is  a solid  of  revolution,  and  the  vertical  projec- 
tions, in  the  two  figures,  are  meridian  sections  of  its 
surface. 

PROBLEM  XVI. 

To  find  the  shades  and  shadows  on  the  shaft  and  base  of  the 
Roman  Doric  column. 

47.  A semi-column  will  illustrate  all  the  cases.  Let 
the  semicircle  AMB  (PI.  10.  Fig.  1)  be  the  horizontal 
projection  of  the  cylindrical  part  of  the  column,  CND 
of  the  upper  and  middle  fillets,  EOF  of  the  upper  torus, 
GPH  of  the  lower  fillet,  LQS  of  the  lower  torus,  and 
the  rectangle  LR  of  the  plinth. 

The  space  included  between  the  semicircle  GPH,  the 
horizontal  projection  of  the  lower  fillet,  and  a semi- 
circle described  at  equal  distances  fromiVMB  and  CND, 
limits  the  horizontal  projection  of  the  scotia.  Let  the 
vertical  projections  be  made  as  in  the  figure. 

Through  the  axis  of  the  column,  let  a meridian  plane 
of  rays  be  drawn — VI  is  its  horizontal  trace.  Through 
I draw  la  at  right  angles  to  IV.  The  plane  of  rays, 
tangent  to  the  cylindrical  part  of  the  shaft,  touches  it 
in  an  element,  which  pierces  the  horizontal  plane  at  a ; 
this  element,  which  is  the  element  of  shade,  can  there- 
fore be  drawn. 

The  element  of  shade  casts  a shadow  on  the  concave 
part  of  the  shaft,  beginning  at  the  point  (a,  a).  This 
shadow  is  in  the  vertical  plane  of  rays  passing  through 
the  element  of  shade,  and  is  therefore  horizontally  pro- 
jected in  the  right  line  drawn  through  a perpendicular 
to  la.  The  vertical  projection  of  this  shadow  is  found 


72 


TREATISE  ON 


by  intersecting  the  shaft  below  d by  horizontal  planes. 
Each  plane  will  intersect  the  shaft  in  a circle,  which 
being  projected  on  the  horizontal  plane,  the  circum- 
ference will  intersect  the  tangent  through  a.  Then,  pro- 
jecting the  point  of  intersection  into  the  vertical  plane, 
we  determine  a point  of  the  vertical  projection  of  the 
shadow. 

The  plane  of  rays  tangent  to  the  upper  fillet,  touches 
it  in  the  element  which  is  projected  on  the  horizontal 
plane  at  b.  This  element  of  shade,  and  the  circum- 
ference of  the  upper  circle  of  the  fillet,  cast  shadows  on 
the  upper  torus.  The  shadow  of  the  element  is  found 
by  a construction  similar  to  that  used  in  finding  the 
shadow  on  the  foot  of  the  shaft.  Points  of  the  shadow 
cast  by  the  circle  are  found  by  first  finding  the  shadow 
of  the  circle  on  a horizontal  plane,  and  then  finding  the 
shadow  which  a horizontal  section  of  the  torus  would 
cast  on  the  same  plane,  and  drawing  rays  through  their 
points  of  intersection.  The  shadow  cast  by  the  circum- 
ference of  the  upper  circle  of  the  fillet  on  the  torus,  is 
made  full,  until  it  intersects  at  (A,  A')  the  curve  of  shade, 
determined  as  in  Prob.  12. 

Passing  the  curve  of  shade  on  the  upper  torus,  the 
construction  for  which  is  given  in  Prob.  12,  we  come 
next  to  the  shadow  which  this  curve  casts  upon  the 
middle  fillet. 

To  find  this  shadow,  we  intersect  the  torus  and  the 
fillet  by  a plane  of  rays  cp',  perpendicular  to  the  vertical 
plane  of  projection.  Through  the  point  c,  in  which 
the  trace  of  this  plane  intersects  the  vertical  projection 
of  the  curve  of  shade,  let  a horizontal  plane  be  passed — 
this  plane  will  intersect  the  torus  in  a horizontal  circle, 
and  let  this  circle  be  projected  on  the  horizontal  plane. 
The  horizontal  projection  of  e is  at  c,  and  c')  is  a 


i ■ 

--Xt 

I 


SHADES  AND  SHADOWS. 


73 


point  of  the  curve  of  shade.  The  ray  of  light  through 
this  point  pierces  the  fillet  in  the  point  (p,  jo'),  which  is 
a point  of  the  curve  of  shadow.  This  shadow  passes 
down  upon  the  fillet  obliquely,  until  it  intersects  the 
lower  circle  of  the  fillet  or  upper  circle  of  the  scotia, 
at  (X,  X').  That  part  of  the  upper  circle  of  the  scotia, 
from  }i  to  (X,  X'),  on  which  the  light  falls,  casts  a shadow 
on  the.  scotia,  which  is  found  as  in  Prob.  13. 

The  ray  of  light  passing  through  the  point  (X,  X'),  in 
which  the  shadow  on  the  fillet  intersects  the  upper  circle 
of  the  scotia,  determines  the  first  point  of  shadow  which 
the  curve  of  shade  on  the  torus  casts  on  the  scotia. 

T o find  points  of  this  shadow,  intersect  the  torus  and 
scotia  by  a plane  of  rays  ef  perpendicular  to  the  verti- 
cal plane ; find  the  point  (e,  e ) in  which  this  plane  inter- 
sects the  curve  of  shade,  and  construct  the  curve  in 
which  it  intersects  the  scotia.  Through  e draw  the  hori- 
zontal projection  of  a ray  of  light — the  point  /,  in  which 
it  intersects  the  curve  of  the  scotia,  is  the  horizontal, 
and  f is  the  vertical  projection  of  a point  of  shadow. 

The  shadow  cast  on  the  scotia  by  the  curve  of  shade 
on  the  torus^  terminates  in  the  upper  circle  of  the  lower 
fillet.  A part  of  this  shadow,  beginning  at  is  seen  in 
horizontal  projection. 

The  element  of  shade  on  the  lower  fillet  is  (g-,  g')  ; 
and  this  element  casts  a shadow,  beginning  at  its  foot, 
on  the  lower  torus.  Then,  that  part  of  the  upper  cir- 
cle of  the  fillet,  intercepted  between  the  element  of  shade 
and  the  point  in  which  it  is  met  by  the  shadow  on  the 
scotia,  casts  a shadow  on  the  lower  torus.  Then  the 
shadow  on  the  lower  torus  is  cast  by  the  curve  of 
shade  on  the  upper  torus ; and  this  shadow  continues 
until  it  intersects  the  curve  of  shade  determined  as  in 
Prob.  12. 


74 


TREATISE  ON 


PROBLEM  XVII. 

To  find  the  shades  and  shadows  on  the  capital  and  shaft  of 
the  Roman  Doric  column, 

48.  The  part  of  the  abacus,  whose  vertical  projec-  , 
tion  is  s r , casts  a shadow  on  the  cyma-reversa. 

The  semi-reversa  is  composed  of  parts  of  cylinders 
whose  elements  are  respectively  parallel  and  perpendi- 
cular to  the  vertical  plane  of  projection.  Two  of  the 
cylinders  intersect  each  other  in-  a curve,  whose  hori- 
zontal projection  is  q”n\  and  vertical  projection  qp7i. 

The  line  (r  V'",  rr  ) of  the'abacus  casts  a line  of  shadow 
on  the  cyma-reversa  parallel  to  itself.  Through  (/•",  r) 
draw  a ray  of  light,  and  find  the  point  in  which  it  pierces 
the  surface  of  the  cylinder  whose  elements  are  parallel 
to  the  vertical  plane : qq  is  the  horizontal  line  drawn 
through  the  point,  and  is  the  line  of  shadow  required. 
The  light  falls  on  a part  of  a cyma-reversa  below  this 
line.  Let  a tangent  plane  of  rays  be  drawn  to  the  part 
of  the  cyma-reversa,  whose  elements  are  parallel  to  the 
vertical  plane.  It  touches  the  cyma-reversa  in  the  ele- 
ment pp,  which  is  therefore  an  element  of  shade.  This 
element  of  shade  casts  a shadow  on  the  part  of  the 
cyma-reversa  below  it.  This  shadow  is  determined  by 
finding  the  intersection  of  the  tangent  plane  of  rays  with 
the  cyma-reversa.  It  is  the  horizontal  line  near  nn. 

The  lower  element  n n of  the  cyma-reversa  casts  a 
shadow  0 d parallel  to  itself,  on  the  part  of  the  abacus 
below  it. 

We  come  next  to  the  echinus,  which  is  the  half  of  a 
torus. 


SHADES  AND  SHADOWS. 


75 


First,  find  the  curve  of  shade  on  the  echinus  as  in 
Prob.  12.  Then,  through  the  horizontal  line  of  the 
abacus  whose  vertical  projection  is  w draw  a plane  of 
rays  : wc  is  its  vertical  trace.  Since  this  plane  is  per- 
pendicular to  the  vertical  plane  of  projection,  the  shadow 
cast  by  the  line  of  the  abacus  is  vertically  projected  in 
the  trace  wc\  The  ray  of  light  through  the  point  w) 
pierces  the  neck  of  the  column  at  the  point  (c,  c ). 

Through  the  lower  line  of  the  abacus  {w'n\  w w)  let 
a plane  of  rays  be  drawn.  This  plane  intersects  the 
echinus  in  a curve  which  is  the  shadow  cast  by  the  line 
{\on\  ww)  on  the  echinus ; this  curve  of  shadow  meets 
the  curve  of  shade  in  the  points  x and  y;  leaving  a part 
of  the  echinus  in  the  light.  The  curve  of  shadow  is 
found  by  intersecting  the  echinus  by  horizontal  planes 
— each  plane  will  intersect  the  echinus  in  a horizontal 
circle  and  the  plane  of  rays  in  a right  line  : the  point  in 
which  the  right  line  intersects  the  circle  is  a point  of  the 
curve  of  shadow 

The  plane  of  rays  passed  through  the  line  (w”n\  w 
intersects  the  neck  of  the  column  in  the  curve  c/,  the 
cavetto  in  the  curve  w,  and  the  fillet  directly  above,  in 
the  curve  v. 

The  curve  of  shade  on  the  echinus  casts  a shadow 
on  the  fillet  below,  which  begins  at  and  intersects  at 
V the  shadow  cast  by  the  lower  line  {w'n\  ww)  of  the 
abacus. 

The  lower  circle  k k'  of  the  fillet,  casts  a shadow  on  the 
echinus,  which  begins  at  a,  and  intersects  the  shadows  cast 
by  the  abacus  in  two  points,  one  near  «,  the  other  at  u. 

F rom  uio  z the  shadow  is  cast  by  the  small  part  of  the 
arc  of  the  circle  kk\  which  is  in  the  light  near  v.  The 
ray  through  z passes  through  the  last  point  of  the  circle 
kk'  which  is  in  the  light,  and  therefore  passes  through  a 
point  of  the  curve  of  shade  on  the  echinus.  From  z to 


76 


TREATISE  ON 


the  line  i i^  the  shadow  is  cast  by  the  curve  of  shade  on 
the  echinus. 

The  lower  circle  ii  of  the  cavetto,  casts  a shadow 
on  the  neck  of  the  column  which  intersects  at  t the 
shadow  cast  by  the  lower  line  {w'n\  w of  the  abacus, 
and  at  t'  the  shadow  cast  by  the  curve  of  shade  on  the 
echinus ; from  ( to  f the  shadow  is  cast  by  the  curve 
of  shade  on  the  echinus. 

We  come  next  to  the  astragal//'.  Find  the  curve 
of  shade  as  in  Prob.  12. 

The  curve  of  shade  on  the  echinus  casts  a shadow 
on  the  astragal  which  intersects  the  curve  of  shade 
near  g.  The  curve  of  shade  on  the  astragal  casts  a 
shadow  on  the  fillet  dd'de. 

The  lower  circle  ee'  of  the  fillet,  casts  a shadow  on 
the  shaft  of  the  column,  which  intersects  the  element  of 
shade  at  6. 

The  shadow  cast  by  a curve  of  shade  on  a surface  of 
revolution,  may  be  found  by  drawing  rays  through  its 
different  points,  and  finding  where  they  pierce  the  sur- 
face ; but  it  is  generally  better  to  pursue  the  method 
adopted  in  Prob.  14. 

OF  THE  HELICOID.* 

49.  The  helicoid  is  a surface  generated  by  a right  line 
moving  uniformly  in  the  direction  of  another  right  line 
which  it  intersects,  and  having  at  the  same  time  a uni- 
form angular  motion  around  it. 

The  fixed  line  is  called  the  axis  of  the  helicoid,  and 
the  moving  line  the  generatrix. 


* The  properties  of  the  helicoid  are  used  in  the  next  problem,  and 
although  their  discussion  belongs  rather  to  warped  surfaces,  than  to  a 
treatise  on  shades  and  shadows,  yet  it  was  thought  best  to  give  the 
properties  here,  as  the  student  may  not  meet  with  them  elsewhere. 


SHADES  AND  SHADOWS. 


77 


Let  (A,  BC)  (PI.  11,  Fig.  n)  be  the  axis  of  a helicoid, 
and  (DA,  D'E)  its  generatrix. 

By  the  definition  of  the  surface,  the  generatrix  has 
two  motions ; one  in  the  direction  of  the  axis  (A,  BC), 
the  other  around  the  axis,  both  of  which  are  uniform. 

Suppose  the  generatrix  to  move  around  the  axis  until 
it  is  horizontally  projected  in  AF,  and  that  during  this 
revolution,  it  has  moved  in  the  direction  of  the  axis  a 
distance  equal  to  EG. 

Through  F draw  FPF'  perpendicular  to  the  ground 
line,  and  make  PF'  equal  to  EG — then  F'G  will  be  the 
vertical  projection  of  the  generatrix,  from  its  new  posi- 
tion; for  every  point  of  the  generatrix  has  moved  an 
equal  distance  in  the  direction  of  the  axis. 

By  considering  the  circumstances  of  the  motion,  it  is 
plain : 

1st.  That  every  point  of  the  generatrix,  except  the 
one  in  which  it  intersects  the  axis,  describes  a curve — 
the  curve  so  described  is  called  a helix. 

2d.  That  since  the  generatrix  preserves  the  same 
position  with  the  axis,  all  the  points  of  the  same  helix 
will  be  equally  distant  from  the  axis,  and  consequently, 
the  projection  of  any  helix  on  a plane  perpendicular  to  the 
axis,  is  in  the  circumference  of  a circle  described  about 
the  centre  A.  The  helix  described  by  the  point  (D,  D') 
is  horizontally  projected  in  the  semicircumference  DIF. 

When  the  generatrix  has  moved  from  one  position  to 
another,  the  motion  in  the  direction  of  the  axis  is  pro- 
portional to  the  part  of  the  axis  which  the  point  of  inter- 
section has  passed  over ; and  all  the  points  of  the  gene- 
ratrix have  moved  equally  in  the  same  direction.  The 
motion  round  the  axis,  or  the  angular  motion  of  the 
generatrix,  is  proportional  to  the  angle  included  between 
two  planes  passing  through  the  axis  and  the  two  posi- 


78 


TREATISE  ON 


tions  of  the  generatrix.  The  angle  between  these 
planes  is  measured  by  the  arc  of  any  circle  described 
about  the  centre  A.  It  may  therefore  be  measured  on 
the  arc  DIFR,  or  on  the  projection  of  any  helix  of  the 
surface. 

Since  the  motion  in  the  direction  of  the  axis,  and  the 
angular  motion  are  both  uniform,  their  measures  will  be 
proportional  to  each  other : that  is,  when  the  point  E 
of  the  generatrix  has  passed  over  a part  of  CB,  the 
angular  motion  of  the  generatrix  is  measured  by  a like 
part  of  DIF. 

Let  the  revolution  of  the  generatrix  (AF,  GF')  be 
continued  until  the  point  (F,  F')  shall  be  horizontally 
projected  at  D.  The  point  G will  then  have  moved  to 
C,  and  the  point  (F,  F')  to  (D,  D")  : GC  and  KD"  being 
each  equal  to  EG.  The  revolution  of  the  generatrix 
may  be  continued  at  pleasure. 

The  horizontal  projection  of  the  helix  described  by 
the  point  (D,  D)  is  the  circumference  DIFR.  It  is 
required  to  find  its  vertical  projection. 

Let  the  semicircumference  DIF  be  divided  into  any 
number  of  equal  parts,  say  eight ; and  divide  the  dis- 
tance D'K  into  the  same  number  of  equal  parts,  and 
draw  the  parallel  lines  as  in  the  figure. 

Now,'  when  the  point  (D,  D')  is  horizontally  project- 
ed at  cf,  it  will  have  ascended  one-eighth  of  the  distance 
D'K,  and  is,  therefore,  vertically  projected  at  a\  Hence 
d is  a point  in  the  vertical  projection  of  the  helix. 
Each  of  the  other  points  of  division  also  gives  a point, 
and  the  vertical  projection  of  the  helix,  from  D'  to  F', 
can  be  drawn ; the  part  F'D"  is  easily  found,  since  it 
has  the  same  position  with  the  horizontal  plane  KF', 
as  the  part  D'a  F'  has  with  the  plane  D'P.  The  part 


SHADES  AND  SHADOWS.  79 

D''X  lias  also  the  same  position  with  the  horizontal 
plane  D"E,  as  the  part  D a'F'  has  with  the  plane  DT. 

The  curve  DQN,  in  which  the  horizontal  plane  of 
projection  intersects  the  surface,  is  determined  by  find- 
ing the  points  in  which  the  elements  of  the  surface 
pierce  the  plane.  This  curve  is  the  spiral  of  Archi- 
medes— it  extends  indefinitely,  unless  we  suppose  the 
revolutions  of  the  generatrix  to  be  limited. 

■ If  through  the  point  (D,  D'),  the  line  (DA,  D'T)  be 
drawn,  making  an  angle  with  the  axis  of  the  surface  equal 
to  the  angle  D'EB,  we  may  regard  this  line  as  the  genera- 
trix of  a helicoid,  similar  in  every  respect  to  the  one 
already  described,  excepting  that  it  has  a different  posi- 
tion with  the  horizontal  plane.  We  shall,  for  the  sake 
of  distinction,  call  the  first  the  upper  helicoid^  and  the 
second  the  lower  helicoid.  The  helix  described  by  the 
point  (D,  D')  is  common  to  the  two  surfaces.  Regard- 
ing the  generatrices  as  indefinite,  the  surfaces  are  inde- 
finite helicoids. 

The  helicoid  is  a warped  surface,  since  the  consecu 
tive  positions  of  the  generatrix  are  not  in  the  same 
plane  (Des.  Geom.  72). 

' It  is  now  required  to  draw  a tangent  plane  to  the  lower  heli- 
coid, at  a point  of  the  common  helix. 

50.  Let  (D,  D')  be  the  point  at  which  the  tangent 
plane  is  to  be  drawn ; and  let  us  find  its  trace  on  the 
horizontal  plane  S'F',  drawn  at  a distance  above  the 
point  of  contact  (D,  D'),  equal  to  the  ascent  of  the 
generatrix  in  half  a revolution. 

The  tangent  plane  must  contain  the  element  (DA, 
D'T),  (Des.  Geom.  89) ; and  therefore,  (S,  S')  where  it 
pierces  the  plane  S'F'  is  one  point  of  the  trace.  But  the 
tangent  plane  must  also  contain  the  tangent  line  to  the 


80 


TREATISE  ON 


helix  at  the  point  (D,  D')  (Des.  Geom.  88)  ; and  these 
two  lines  determine  its  position. 

The  helix  (DaF,  D'a  F')  is  on  the  surface  of  the  right 
cylinder  which  projects  it  on  the  horizontal  plane.  Con- 
ceive a tangent  plane  to  be  drawn  to  this  cylinder  along 
the  element  which  pierces  the  horizontal  plane  at  D. 
The  tangent  line  to  the  helix  at  (D,  D')  will, be  con- 
tained in  this  plane. 

Let  the  surface  of  the  right  cylinder  be  developed  on 
the  tangent  plane.  The  semicircumference  DIF  will 
develop  into  a right  line  (Des.  Geom.  131)  ; and  since 
the  arcs  D«,  D^,  &c,  are  proportional  to  the  dis- 
tances of  the  corresponding  points  of  the  helix  above 
the  horizontal  plane,  it  follows  that  the  helix  will  de- 
velop into  a right  line. 

Then,  lay  off  DV  equal  to  the  semicircumference 
DIF,  and  draw  the  perpendicular  VF"',  and  make  it  equal 
to  PF' : we  have  then  DF"  for  the  development  of  the 
helix.  But  the  tangent  line  at  (D,  D')  and  the  helix 
make  the  same  angle  with  the  horizontal  plane  ; there- 
fore, the  helix  developed  coincides  with  the  tangent  line 
at  (D,  D').  Hence,  the  tangent  line  to  the  helix  at  the 
point  (D,  D'),  pierces  the  horizontal  plane  S'F'  at  the 
point  (V,  K),  a distance  from  the  plane  DA  equal  to  the 
semicircumference  DIF.  Therefore,  SV  is  the  hori- 
zontal projection  of  the  trace  of  the  tangent  plane 
on  the  plane  S'F'. 

51.  It  is  evident,  from  the  similarity  of  triangles,  that 
if  the  horizontal  plane  on  which  the  trace  is  found,  be 
taken  above  or  below  ST',  the  distance  D'K  will  be  to  the 
semicircumference  DIF,  as  the  distance  from  D'  to  the 
plane,  is  to  the  distance  from  the  vertical  plane  DAF  to 
where  the  tangent  pierces  the  horizontal  plane.  There- 
fore, if  the  horizontal  plane  ST'  were  passed  through  D", 


SHADES  AND  SHADOWS. 


81 


making  the  distance  from  D'  equal  to  the  ascent  of  the 
generatrix  during  an  entire  revolution,  the  distance  DV 
would  be  equal  to  the  entire  circumference  DIFR. 

52,  By  considering  the  generation  of  the  helicoid,  it 
is  evident  that  each  helix  makes  a different  angle  with 
the  horizontal  plane ; the  limits  of  these  angles  being 
90°  and  0. 

For,  the  helix  described  by  the  point  in  which  the 
generatrix  mtersects  the  axis,  is  the  axis  itself,  and  is 
therefore  perpendicular  to  the  horizontal  plane.  The 
helix  described  by  the  point  at  an  infinite  distance  from 
tho  axis  may  be  considered  horizontal,  since  the  dis- 
tance which  the  point  moves  in  the  direction  of  the 
axis  is  inconsiderable  in  comparison  with  its  angular 
motion. 

Since  every  tangent  plane  to  the  helicoid  passes 
through  an  element  (Des.  Geom.  89),  the  limits  of  the 
angles  which  tangent  planes  make  with  the  horizontal 
plane  are  90°,  and  the  angle  made  by  the  generatrix, 

PROBLEM  XIX. 

Having  given  the  projections  of  a screw  and  the  direction 
of  the  lights  it  is  required  to  find  the  shades  and  shadows  on 
the  dijferent  parts  of  the  screw ^ and  the  shadow  cast  on  a hori-- 
zontal  plane. 

53.  Let  there  be  a vertical  cylinder  whose  axis,  is 
(A,  A'B),  (PI.  11)  and  the  radius  of  whose  base  is  AC. 

Let  the  base  of  an  isosceles  triangle,  whose  vertical 
projection  is  CD'E,  be  placed  to  coincide  with  the  ele- 
ment of  the  cylinder  which  pierces  the  horizontal  plane 
at  C ; the  vertex  D'  of  the  triangle  being  in  the  hori- 
zontal plane  at  D.  The  plane  of  the  triangle  will  pass 

F 


82 


TREATISE  ON 


through  the  axis  of  the  cylinder,  and  therefore  the  two 
equal  sides  will  both  intersect  it. 

Let  the  triangle  be  now  revolved  about  the  cylinder, 
having,  at  the  same  time,  a uniform  motion  in  the  direc- 
tion of  the  elements — the  plane  of  the  triangle  con- 
tinuing to  pass  through  the  axis. 

When  the  triangle  has  been  revolved  half  round  the 
cylinder,  suppose  it  to  have  ascended  a distance  equal 
to  half  its  base.  The  vertex  (D,  D')  will  then  have  the 
position  (G,G') ; G"G'  being  equal  to  half  the  base. 
After  the  remaining  half  of  the  revolution  is  completed, 
the  vertex  of  the  triangle  will  have  ascended  a distance 
equal  to  its  base,  and  will  be  vertically  projected  at  F; 
FEH  will  then  be  the  vertical  projection  of  the  gene- 
rating triangle. 

The  cylinder  about  which  the  triangle  has  been  re- 
volved, is  called  the  cylinder  of  the  screiv. 

The  solid  generated  by  the  triangle,  in  an  entire  revo- 
lution, is  called  the  thread  of  the  screw.  The  surface 
generated  by  the  upper  equal  side  of  the  triangle  is 
called  the  upper  surface  of  the  thread  ; and  the  surface 
generated  by  the  other  equal  side  is  called  the  lower 
surface  of  the  thread.  These  surfaces  are  portions  of 
helicoids  (49).  The  helicoids  may  be  considered  inde- 
finite, by  supposing  the  equal  sides  of  the  triangle  to  be 
indefinitely  produced. 

The  helix  described  by  the  vertex  (D,  D'),  is  called 
the  outer  helix  of  the  screw ; and  the  one  described  by 
either  extremity  of  the  base,  is  called  the  inner  helix 
The  curves  described  by  the  intermediate  points  are 
called  helices. 

The  vertical  projections  of  the  outer  and  inner  helices 
are  constructed  as  in  Art  49.  The  parts  which  lie  in 
front  of  the  plane  DAG  are  made  full,  those  which  lie 


SHADES  AND  SHADOWS. 


83 


behind  it  are  dotted.  The  horizontal  projection  of  the 
outer  helix  is  the  circumference  described  with  the 
radius  AD,  and  of  the  inner  helix,  the  circumference 
described  with  the  radius.  AC. 

The  outer  helix  rises  from  the  horizontal  plane  at  D, 
and  passes  behind  the  plane  DAG  at  (G,  G').  The 
inner  helix  rises  above  the  horizontal  plane  at  I,  and 
passes  in  front  of  the  plane  DAG  at  (C^  E). 

If  the  triangle  EFH  be  again  revolved  around  the 
cylinder  of  the  screw,  it  will  generate  a second  thread ; 
and  every  new  revolution  will  give  an  additional  thread. 

The  base  of  the  triangle,  or  the  distance  which  it  has 
ascended  in  an  entire  revolution,  is  called  the  distance 
between  the  threads^ 

The  curve  into  which  the  surface  of  the  screw  inter- 
sects the  horizontal  plane  is  found  as  in  Art.  49,  except- 
ing that  the  elements  are  differently  inclined  to  the  hori- 
zontal plane,  which,  however  does  not  vary  the  prin- 
ciples of  the  construction.-^It  is  the  curve  D^^^I. 

The  screw  is  usually  worked  into  a block  of  wood 
called  a fillet.  W e have  taken  the  fillet  octagonal ; the 
octagon  4 1 2 3 5 is  its  horizontal  projection,  and  the 
rectangle  above  the  threads  its  vertical  projection. 

It  is  now  required  to  find  the  curve  of  shade  on  the 
lower  surfaces  of  the  threads. 

Let  a plane  be  drawn  tangent  to  the  lower  surface  of 
the  thread,  at  the  point  (D,  F)  (50),  and  let  its  trace  be 
constructed  on  the  horizontal  plane  L'H,  taken  at  a 
distance  above  the  point  of  contact  equal  to  the  half 
distance  of  the  threads. 

The  point  (L,  L'),  in  which  the  element  passing 
through  the  point  of  contact  pierces  the  horizontal 
plane  L'H,  is  one  point  of  the  trace  of  the  tangent 
plane.  Drawing  through  D the  perpendicular  DO,  and 

F 2 


84 


TREATISE  ON 


making  it  equal  to  the  semicircumference  DaG,  we  have 
the  distance  from  the  plane  DAG,  at  which  the  line 
tangent  to  the  helix  at  (D,  F)  pierces  the  horizontal 
plane  L'H  (50).  This  distance  should  have  been  laid 
off  in  front  of  the  plane  DAG,  but  this  could  not  be 
done,  for  want  of  room  on  the  paper.  If  however,  we 
suppose  a line  drawn  through  L and  the  point  O,  taken 
at  a distance  equal  to  DO  in  front  of  the  plane  DAG,  it 
will  make  an  angle  with  LD  equal  to  the  angle  DLO. 
Hence,  if  MJc  be  drawn,  making  the  angle  DL^'  equal 
to  the  angle  DLO,  it  will  be  the  horizontal  projection  of 
the  trace  of  the  tangent  plane.  The  tangent  plane  cuts 
the  axis  of  the  screw  at  (A,  I '). 

Now,  if  this  plane  be  a plane  of  rays,  the  point  of 
contact  (D,F)  will  be  a point  of  the  curve  of  shade. 

To  ascertain  if  it  be  a plane  of  rays,  draw  through 
(A,  T)  a ray  of  light;  its  projections  are  AK  and  I'K'. 

If  the  tangent  plane  be  a plane  of  rays,  the  ray  of 
light  having  one  point  in  common,  will  coincide  with  it ; 
and  if  it  coincide  with  it,  it  will  pierce  the  horizontal 
plane  LTI  in  the  trace  MJc,  But  it  pierces  this  horizon- 
tal plane  at  (K,  K')  out  of  the  trace — -therefore,  the  . 
tangent  plane  is  not  a plane  of  rays. 

Let  the  ray  (AK,  I'K)  be  revolved  about  the  axis  of 
the  screw ; it  will  generate  the  surface  of  a right  cone 
whose  vertex  is  (A,!'),  and  the  point  (K,  K')  will  describe, 
in  the  horizontal  plane  K'LH,  the  arc  of  a circle  of 
which  kKM  is  the  horizontal  projection.  The  tangent 
plane  to  the  screw  will  intersect  the  surface  of  this  cone 
in  two  elements  whose  horizontal  projections  are 
Ak  and  Ak\  - 

These  lines  may  be  regarded  as  the  ray  (AK,  I'K') 
after  it  has  been  revolved  about  the  axis  of  the  screw 
to  coincide  with  the  tangent  plane. 


SHADES  AND  SHADOWS. 


85 


Let  the  tangent  plane  be  now  revolved  around  the 
axis  of  the  screw,  in  such  a manner  as  to  continue  tan- 
gent along  the  outer  helix,  until  the  revolved  ray  Ah 
shall  be  horizontally  projected  in  AK.  In  this  revolution 
of  the  tangent  plane,  every  point  of  the  line  Ah  will 
ascend  equally ; therefore,  after  the  counter  revolution 
the  vertical  projection  of  the  line  will  be  parallel  to  FK' ; 
hence,  the  line  itself  will  be  a ray  of  light. 

In  the  revolution  of  the  tangent  plane,  every  point 
has  an  equal  angular  motion,  and  this  motion  is  mea- 
sured by  the  arc  oDb ; the  point  of  contact  (D,  F)  having 
the  same  angular  motion,  if  we  lay  off  T>ha  equal  to 
oD6,  or  what  is  the  same,  ba  equal  to  oD,  we  find  a,  the 
horizontal  projection  of  the  point  of  contact  when  the 
tangent  plane  is  a plane  of  rays.  Projecting  a into  the 
vertical  projection  of  the  outer  helix,  we  have  one  point 
(a,  a ) of  the  curve  of  shade. 

If  v/e  revolve  the  tangent  plane  from  D towards  o,  as 
we  are  at  liberty  to  do,  causing  it  to  descend  along  the 
outer  helix  but  continuing  tangent  to  the  surface,  until 
the  line  Ah'  shall  be  horizontally  projected  in  AK,  the 
plane  will,  for  the  reasons  given  above,  become  a plane  of 
rays,  and  the  point  of  contact,  a point  of  the  curve  of 
shade.  The  angular  motion,  in  this  revolution,  will  be 
measured  by  be — hence,  laying  off  Dc  equal  to  bc^  we 
have  c,  the  horizontal  projection  of  the  point  of  contact. 
Projecting  the  point  c into  the  vertical  projection  of  the 
outer  helix  at  c",  determines  (c,  c'),  a point  in  a second 
curve  of  shade.  Hence  we  see  that  two  tangent  planes 
of  rays  can  be  drawn  touching  the  same  thread  of  the 
screw  in  points  of  the  outer  helix. 

Let  a tangent  plane  be  drawn  to  the  under  surface  of 
the  thread  of  the  screw,  at  the  point  (C,  E)  of  the  inner 
helix ; and  let  its  trace  be  constructed  on  the  horizontal 


86 


TREATISE  ON 


plane  KX'H,  at  a distance  above  the  point  of  contact 
equal  to  the  ascent  of  the  generating  triangle  in  an  entire 
re  volution.  The  point  (L,  U),  in  which  the  element 
through  the  point  of  contact  pierces  the  horizontal  plane 
UH,  is  one  point  of  the  trace,  and  making  CP  equal  to 
the  circumference  Ceflf  (51),  and  considering  P in  front 
of  the  plane  DAG,  we  have  a secondpoint  of  the  trace,  it 
being  the  point  in  which  the  tangent  line  to  the  helix  at 
(C,  E)  pierces  the  plane  L'H.  Drawing  LP,  and  then 
the  line  Lc?',  making  the  angle  DLc^'  equal  to  the  angle 
DLP,  we  determine  d'l^d  the  trace,  on  the  plane  UH,  of 
the  tangent  plane  to  the  inner  helix  at  the  point  (C,  E). 

This  tangent  plane  cuts  the  axis  of  the  screw  at  the 
point  (A,  F),  and  since  the  ray  through  this  point  does 
not  pierce  the  horizontal  plane  K'H  in  the  trace  of  the 
tangent  plane,  it  follows,  that  the  tangent  plane  is  not  a 
plane  of  rays.  Let  the  ray  (AK,  FK')  be  revolved  about 
the  axis  of  the  screw  as  before-r-it  will  generate  the  sur- 
face of  a right  cone,  which  the  tangent  plane  will  inter- 
sect in  two  elements  whose  horizontal  projections  are 
Kd  and  Ad'. 

The  tangent  plane  is  now  revolved  to  become  a plane 
of  rays,  as  in  the  last  case.  The  angular  motion,  when 
it  ascends,  is  eC^,  and  when  it  descends,  fg-  There- 
fore, making  ge  equal  to  Ce,  and  Cf  equal  to  gf^  we  have 
e and  f the  horizontal  projections  of  the  two  points  of 
tangency.  The  point  e is  vertically  projected  on  the 
inner  helix  at  e',  and  the  point  f at  /"—hence,  two 
points  in  each  of  the  curves  of  shade  are  found.  It  is 
now  required  to  find  intermediate  points. 

Any  element  of  the  lower  surface  of  the  thread,  whose 
horizontal  projection  passes  between  the  points  a and  c, 
will  contain  an  intermediate  point  of  the  curve  of  shade. 


SHADES  AND  SHADOWS. 


87 


Assume  Ah  for  the  horizontal  projection  of  such  an 
element. 

Projecting  the  point  h into  the  outer  helix  at  /i',  and 
the  point  ^ into  the  inner  helix  at  e',  we  find  R^7l'  the 
vertical  projection  of  the  element. 

Through  Qi^  the  upper  extremity  of  the  element, 
draw  the  horizontal  plane  M'/i',  and  through  the  point 
(A,  R)  draw  a ray  of  light  (AM,  RM') ; this  ray  pierces 
the  horizontal  plane  at  the  point  (M,  M'). 

If  through  the  element  (A/i,  R/i')  we  suppose  a plane 
of  rays  to  be  passed,  its  trace  on  the  plane  MK  will  be 
the  line  of  which  MA  is  the  horizontal  projection;  and 
since  the  surface  of  the  screw  is  a warped  surface,  this 
plane  will  be  tangent  to  it  at  some  point  of  the  element 
(Des.  Geom.  229) : the  point  of  contact  is  a point  of 
the  curve  of  shade. 

Let  us  suppose  for  a moment  this  point  of  contact  to 
be  found,  and  that  it  is  (m,  At  a distance  above  the 
point  of  contact  equal  to  half  the  distance  of  the 
threads,  let  the  horizontal  plane  N'H'  be  drawn.  The 
trace  of  the  plane  of  rays  on  this  plane  is  parallel  to  its 
trace  on  the  plane  MK ; therefore,  NA"  drawn  parallel 
to  MA  is  the  trace  on  the  plane  N'H'. 

If  then,  through  the  point  w,  a perpendicular  be  drawn 
to  Aw,  and  produced  till  it  meets  NA"  in  Q,  the  distance 
wQ  will  be  the  semicircumference  of  the  circle  whose 
radius  is  Aw  (50).  But  A' w is  the  horizontal  projection  of 
a part  of  the  element  intercepted  between  two  horizontal 
planes,  at  a distance  from  each  other  equal  to  the  half 
distance  of  the  threads  ; and  since  all  the  elements  make 
the  same  angle  with  the  horizontal  plane,  this  projection 
is  equal  to  the  projection  of  D'E,  that  is,  to  DC  or  hu 
Hence,  if  through  2,  ^S  be  drawn  perpendicular  to  A?7i, 
we  shall  have  two  similar  triangles.  A' wQ  and  AtS,  having 


88 


TREATISE  ON 


in  each  an  equal  homologous  side  h!'  u and  hi ; hence  the 
sides  wQ,  and  ^S  are  equal.  But  the  line  wQ  has  been 
proved  equal  to  the  semicircumference  of  the  circle 
passing  through  the  point  of  contact ; therefore,  iS  is 
also  equal  to  this  semicircumference. 

After  having  drawn  the  trace  M/i,  of  the  plane  of  rays, 
we  have  only  to  draw  iS  perpendicular  to  Aih  and  find  the 
radius  of  the  circle  of  which  ^S  is  the  semicircumference. 
This  is  easiest  done  by  laying  off  iS  from  D,  on  DO,  and 
drawing  through  the  extremity  of  the  line  a parallel  to 
the  line  joining  A and  O — the  distance  cut  off  from  D,  on 
DA,  will  be  the  radius  required ; for  the  semicircumfer- 
ences are  to  each  other  as  their  radii.  Then,  with  this 
radius,  and  A as  a centre,  describe  an  arc ; the  point  w, 
where  it  cuts  Ah,  is  the  horizontal  projection  of  the 
point  of  contact,  and  this  point  being  projected  into  the 
vertical  projection  of  the  element  atw,  the  vertical  pro^ 
jection  of  the  point  of  contact  is  also  determined.  An 
intermediate  point  in  the  curve  of  shade  on  the  other 
side  of  the  plane  DAG,  may  be  found  by  a similar  con- 
struction. 

Since  the  light  has^  the  same  position  with  the  lower 
surfaces  of  all  the  threads,  the  curves  of  shade  upon 
them  will  be  directly  over  each  other : therefore,  if  a 
vertical  line  be  drawn  through  (a,  a),  the  points  in  which 
it  cuts  the  outer  helices  will  be  points  of  the  curves  of 
shade,  and  the  vertical  line  through  (e , e'),  will  deter- 
mine corresponding  points  of  shade  on  the  inner  helices. 

It  is  now  required  to  find  the  shadow  which  any  thread 
will  cast  on  the  surface  of  the  thread  directly  below  it. 
The  curve  of  shade,  and  a part  of  the  outer  helix, 
will  cast  shadows  on  the  thread  below  them.  The  sha- 
dow will  begin  at  the  point  where  the  curve  of  shad© 
meets  the  inner  helix. 


SHADES  AND  SHADOWS. 


89 


To  find  the  shadow  cast  by  the  curve  of  shade,  we 
first  find  the  shadow  which  the  curve  of  shade  would 
cast  on  the  horizontal  plane  if  the  screw  were  removed. 
This  is  done  by  drawing  rays  of  light  through  its  seve- 
ral points,  and  finding  where  they  pierce  the  horizontal 
plane.  We  then  take  any  element  of  the  upper  surface, 
near  to  (e\  c"),  on  which  we  suppose  a shadow  will  fall, 
and  find  the  shadow  which  it  would  cast  on  the  horizon- 
tal plane.  Through  the  point  where  these  shadows 
intersect,  if  we  suppose  a ray  of  light  to  be  drawn,  it 
will  intersect  both  the  element  and  the  curve  of  shade. 
The  point  where , it  meets  the  element  is  the  point  of 
shadow  on  the  element,  and  the  point  where  it  meets  the 
curve  of  shade  is  the  point  casting  the  shadow.  The 
ray  through  p gives  {p\p')  for  the  point  of  shadow. 

To  find  the  shadow  cast  by  the  helix.  Take  any  ele- 
ment on  which  we  suppose  the  shadow  will  fall,  and 
through  it  pass  a plane  of  rays.  Find  the  point  in  which 
this  plane  cuts  the  outer  helix,  directly  above,  and 
through  this  point  draw  a ray  of  light:  where  it  inter- 
sects the  element  is  a point  of  the  required  shadow. 

Let  Aq  be  the  horizontal  projection  of  an  element  on 
which  it  is  supposed  the  sliadow  will  fall,  and^^'t;  its 
vertical  projection.  Through  (g,  q),  the  foot  of  the  ele- 
ment, let  the  horizontal  plane  q'T'  be  drawn.  Through 
(A,  v),  the  point  in  which  the  element  intersects  the  axis, 
let  a ray  of  light  be  drawn this  ray  pierces  the  horizon- 
tal plane  through  (5^,  q)  at  (T,  T')-— therefore  ^T  is  the 
trace,  on  the  plane  ^'T',  of  the  plane  of  rays  through 
the  element.  It  is  now  required  to  find  the  point  in 
which  this  plane  of  rays  cuts  the  outer  helix  (DagrG,  FX.) 

Through  the  axis  of  the  screw  let  a plane  be 
passed  perpendicular  to  the  plane  of  rays — its  hori- 
zontal trace  is  Ar  perpendicular  to  qT.  Let  the  plane 


90 


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of  rays  and  the  plane  whose  horizontal  trace  is  Ar  be 
revolved  about  the  axis  of  the  screw  until  Ar  becomes 
parallel  to  the  ground  line ; and  suppose  the  plane  of 
rays  to  have  at  the  sanie  time  an  ascending  motion  in 
the  direction  of  the  axis,  such  that  (A^^,  vq)  shall  con- 
tinue an  element  of  the  screw.  When  Ar  is  parallel  to 
the  ground  line,  the  plane  of  rays  will  be  perpendicular 
to  the  vertical  plane  of  projection,  and  the  angular 
motion  will  be  measured  by  the  arc  rG. 

From  q laying  oif  qq"  equal  to  rG:  Aq"  will  be  the 
horizontal  projection  of  the  element  after  the  revolution, 
and  q"s  will  be  its  vertical  projection.  But,  since  the 
element  is  a line  of  the  plane  of  rays,  which,  in  its  re- 
volved position  is  perpendicular  to  the  vertical  plane, 
the  projection  of  the  element  is  the  vertical  trace  of  the 
plane  (Des.  Geom.  24). 

Therefore  /,  the  point  in  which  the  line  q”s  inter- 
sects FX,  is  the  vertical  projection  of  the  point  in  its 
revolved  position,  in  which  the  plane  of  rays  cuts  the 
outer  helix.  The  horizontal  projection  of  this  point  is 
at  s.  In  the  counter  revolution,  the  point  (5, s)  has  the 
same  angular  motion  as  the  other  points  of  the  element 
— therefore,  laying  off  ss”  equal  to  rG,  and  projecting  s" 
into  the  outer  helix  at  s'\  determines  (s'\  s")  the  point 
in  which  the  plane  of  rays  cuts  the  outer  helix. 

Through  this  point  draw  a ray  of  light — the  point 
(n,  n ),  where  it  intersects  the  element,  is  a point  of  the 
curve  of  shadow.  By  similar  constructions  we  may  find 
other  points  of  the  curve.  Having  found  several  points 
of  the  curve  of  shadow,  let  its  two  projections  e'p'nt^ 
e''p"n't'  be  drawn. 

The  shadows  on  the  threads  directly  above,  are 
similar  to  the  one  already  found,  and  are  all  horizon- 
tally projected  in  the  curve  e'p'n  t. 


SHADES  AND  SHADOWS. 


91 


Drawing  through  t'  the  vertical  projection  of  a ray 
of  light,  the  point  /,  where  it  meets  the  outer  helix  above, 
is  the  vertical  projection  of  the  point  which  casts  the 
shadow  at  ; and  a'l  is  the'vertical  projection  of 
the  part  of  the  outer  helix  which  casts  a shadow  on  the 
thread. 

There  are  similar  shadows  on  the  other  side  of  the 
plane  of  rays  KAT,  which  are  constructed  in  the  same 
manner  as  those  already  found. 

If  through  any  element  of  the  upper  surface  of  a 
thread,  a plane  of  rays  be  drawn,  and  its  trace  con- 
structed on  the  horizontal  plane  through  the  foot  of 
the  element,  the  horizontal  projections  of  these  traces 
will  all  pass  through  the  point  T. 

F or,  suppose  we  take  the  element  whose  vertical  pro- 
jection is  q"'s'.  Through  q'"  draw  the  horizontal  plane 
q'''T",  The  ascent  of  the  foot  of  this  element, in  revolving 
from  the  position  (Ag^,  ?;5^'),is  equal  to  T'T",  and  the  point 
V has  ascended  the  same  distance.  Therefore  the  ver- 
tical projection  of  the  ray  drawn  through  the  point  in 
which  the  element  intersects  the  axis,  will  intersect  the 
vertical  line  T'T"  at  T",  which  is  in  the  horizontal  plane 
^'"T" ; and  as  the  same  may  be  shown  for  all  other 
elements,  it  follows,  that  the  horizontal  projections  of 
all  the  traces  will  pass  through  the  point  T. 

To  find  the  shadow  cast  on  the  screw  by  the  fillet : 

The  lower  lines  of  the  fillet  which  are  towards  the 
source  of  light,  and  whose  horizontal  projections  are 
12,  23,  and  14,  will  cast  shadows  on  the  screw. 

Let  it  be  required  to  find  the  shadow  cast  by  the  line 
whose  horizontal  projection  is  12. 

Through  this  line  suppose  a plane  of  rays  to  be  passed 
-^the  curve  in  which  it  intersects  the  surface  of  the 
screw  is  the  curve  of  shadow  required.  The  plane  of 


92 


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rays  through  the  axis  of  the  screw,  cuts  the  lower  line 
12  of  the  fillet  at  the  point  (y,  y'y  Through  y draw  the 
vertical  projection  of  a ray — the  point  y\  where  it  inter- 
sects the  axis  of  the  screw,  is  the  point  in  which  the 
plane  through  12  cuts  the  axis. 

Every  plane  passing  through  the  axis  of  the  screw  will 
intersect  the  surface  in  an  element,  and  the  plane  of  rays 
in  a right  line  passing  through  the  point  (A,  y"^ ; the 
point  in  which  these  lines  intersect  is  a point  of  the 
shadow. 

The  plane  whose  trace  is  KA  intersects  the  thread  on 
which  the  shadow  falls  in  the  element  {bg^  and  the 
plane  of  rays  in  the  line  whose  vertical  projection  yy" ; 
hence  their  point  of  intersection  is  a point  of  the  curve 
of  shadow.  The  plane  LA  gives  the  point  of  shadow 
on  the  element  parallel  to  the  vertical  plane. 

The  plane  of  rays  through  23,  cuts  the  axis  of  the 
screw  at  X".  Having  found  a sufficient  number  of  points, 
let  the  vertical  projection  of  the  curve  be  drawn.  The 
shadow  passes  off  the  thread  at  m.  The  horizontal  pro- 
jection is  easily  found,  but  is  not  made,  lest  it  should 
render  the  figure  on  the  horizontal  plane  too  complicated. 

To  find  the  shadow  cast  by  the  screw  on  the  hori- 
zontal plane  : 

This  shadow  lies  on  both  sides  of  the  plane  KA ; but 
as  the  construction  for  the  two  parts  is  the  same,  it  will 
only  be  necessary  to  make  it  for  one  of  them. 

The  shadow  on  the  horizontal  plane  begins  at  the 
point  where  the  curve  of  shade  on  the  thread  inter- 
sects the  horizontal  plane.  The  curve  of  shade  between 
this  point  and  the  outer  helix,  casts  a shadow  on  the  hori 
zontal  plane;  this  shadow  is  found  by  drawing  rays 
through  the  points  of  shade,  and  finding  where  they 
pierce  the  horizontal  plane. 


SHADES  AND  SHADOWS. 


93 


The  part  of  the  outer  helix,  between  the  points  a and 

next  casts  a shadow.  This  shadow  is  found  by  draw- 
ing rays  through  the  points  of  the  helix,  and  finding 
where  they  pierce  the  horizontal  plane.  The  point  (/,  /') 
casts  its  shadow  at  t'\  and  zt”  the  shadow  cast  by  the 
curve  of  shade  and  outer  helix. 

The  ray  through  (/,  /')  intersects  the  outer  helix  of 
the  next  higher  thread  at  L The  point  whose  vertical 
projection  is  I casts  a shadow  on  the  screw  at  (/,  ^'),  and 
on  the  horizontal  plane  at  t'\  The  part  of  the  helix 
from  I to  the  point  where  it  intersects  the  shadow  on  the 
thread,  will  cast  a shadow  on  the  horizontal  plane,  which 
is  found  by  drawing  rays  through  the  points  of  the  helix, 

We  then  pass  to  the  helix  of  the  next  higher  thread, 
and  so  on  until  we  arrive  at  the  thread  on  which  the 
shadow  of  the  fillet  falls. 

Having  before  found  m,  the  vertical  projection  of  the 
point  in  which  the  shadow  of  the  fillet  intersects  the 
outer  helix,  we  find  the  shadow  of  this  point  on  the  hori- 
zontal plane  which  is  at  m'.  At  this  point  the  shadow 
of  the  fillet  on  the  horizontal  plane  begins.  This 
shadow  is  found  too  easily  to  require  particular  expla- 
nation. 

54.  The  brilliant  point  of  the  surface  of  the  helicoid, 
is  found  by  bisecting  the  angle  between  a ray  of  light 
and  a line  drawn  to  the  eye,  and  then  drawing  a plane 
perpendicular  to  the  bisecting  line  and  tangent  to  the 
surface.  The  details  of  the  construction  are  left  as  an 
exercise  for  the  student. 

55.  In  all  the  constructions  v.diich  have  been  made, 
the  rays  of  light  have  been  supposed  parallel.  It  may 
be  well  to  consider  how  the  constructions  would  have 
been  made  had  the  rays  been  divergent  or  convergent. 

If  we  suppose  the  rays  of  light  to  emanate  from  a 


94 


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luminous  point  at  a finite  distance,  and  to  fall  upon  an 
opaque  body,  the  rays  will  be  divergent. 

Let  us  suppose  a cone  to  be  drawn  tangent  to  the 
opaque  body,  of  which  the  luminous  point  shall  be  the 
vertex.  It  is  plain, 

1®.  That  the  line  of  contact  of  this  cone  with  the 
opaque  body  will  be  the  line  of  shade. 

2®.  That  the  part  of  space  included  within  the  sur- 
face of  this  cone,  and  lying  on  that  side  of  the  opaque 
body  opposite  to  the  source  of  light,  will  be  the  indefi- 
nite shadow  of  the  opaque  body. 

3°.  That  the  shadow  of  the  opaque  body  on  any  sur- 
face, will  be  the  intersection  of  that  surface  with  the 
tangent  cone  of  rays. 

56.  If  we  suppose  the  light  to  emanate  from  a lumin- 
ous body,  we  may  suppose  a cone  drawn  tangent  to  the 
luminous  body,  and  to  any  opaque  body  whose  shade 
and  shadow  are  to  be  determined.  The  curve  of  con- 
tact on  the  opaque  body  will  be  the  line  of  shade,  and  the 
part  of  space  within  the  surface  of  the  tangent  cone 
and  on  the  side  of  the  opaque  body  opposite  the  source 
of  light,  will  be  the  indefinite  shadow.  When  the  opaque 
body  is  the  largest,  the  luminous  body  and  the  vertex 
of  the  tangent  cone  will  be  on  the  same  side  of  the 
opaque  body,  and  the  rays  of  light  will  be  divergent ; but 
when  it  is  the  smallest,  the  opaque  body  will  be  between 
the  luminous  body  and  the  vertex  of  the  cone,  and  the 
rays  of  light  may  be  considered  as  converging  to  the 
vertex  of  the  tangent  cone. 

In  all  the  cases  in  which  the  rays  of  light  are  not 
parallel,  the  problems  in  Shades  and  Shadows  will  be 
solved  by  finding  the  contact  of  a cone  with  an  opaque 
body,  and  the  intersection  of  this  cone  with  any  surface 
on  which  the  shadow  falls. 


LINEAR  PERSPECTIVE. 


CHAPTER  I. 

57.  Had  we  no  knowledge  of  objects  other  than  what 
is  derived  through  the  medium  of  sight,  we  should  sup- 
pose them  to  differ  from  each  other  in  two  respects  only 
— form  and  colour.  Objects  having  different  forms,  or 
different  colours,  produce  different  effects  upon  the 
eye ; but  objects  of  the  same  form  and  colour  cannot  be 
distinguished  from  each  other  without  the  aid  of  the 
other  senses. 

58.  When  we  view  an  object,  all  the  points  of  it  which 
are  seen  are  supposed  either  to  emit  or  reflect  rays  of 
light  which  fall  upon  the  eye ; and  it  is  through  the 
medium  of  these  rays  that  we  derive  the  idea  both  of 
its  form  and  colour. 

59.  Perspective  is  the  art  of  representing  objects  on 
a surface,  in  such  a manner  that  the  representations 
shall  present  to  the  eye,  situated  at  a particular  point, 
the  same  appearance  as  is  presented  by  the  objects 
themselves.  The  representation  of  an  object  so  made 
is  called  its  perspective, 

60.  Let  us  now  suppose  that  we  are  viewing  an  object 
in  space,  and  that  a transparent  plane  is  placed  between 
us  and  the  object. 

Every  point  of  the  object  which  is  seen,  is  supposed 
to  emit  a ray  of  light  that  falls  upon  the  eye,  and  each 


96 


TREATISE  ON 


ray  pierces  the  transparent  plane  in  a point.  If  to  each 
point  so  determined,  a proper  colouring  be  given,  the 
representation  or  picture,  on  the  transparent  plane,  will 
present  to  the  eye  the  same  appearance  as  the  object 
itself.  Such  representation  is,  therefore,  the  perspective 
of  the  object. 

61.  The  rays  of  light  coming  from  the  different 
points  of  the  object  to  the  eye,  are  called  visual  rays ; 
and  the  plane  on  which  the  representation  is  made,  is 
called  the  Perspective  Plane. 

62.  The  art  of  Perspective  is  therefore  divided  into 
two  parts. 

1®.  To  find  the  points  in  which  the  visual  rays  pierce 
the  perspective  plane,  which  determines  the  general 
outline  of  the  perspective. 

2®.  So  to  shade  and  colour  this  outline,  that  it  shall 
appear  in  every  respect  like  the  object  itself. 

The  first  part  is  called  Linear  Perspective.  It  em- 
braces that  portion  of  perspective  that  is  strictly  mathe- 
matical, and  which  will  form  the  subject  of  the  following 
treatise. 

The  second  part  is  called  Aerial  Perspective.  This 
branch  of  the  art  belongs  to  the  draftsman  and  the 
painter,  and  is  to  be  learned  by  a careful  study  of  the 
objects  of  nature,  under  the  guidance  of  an  improved 
and  cultivated  taste. 

63.  As  it  is  the  end  of  perspective  to  represent  objects 
as  they  appear  in  nature,  such  a position  ought  to  be 
given  to  the  perspective  plane  as  will  enable  us  to  con- 
ceive, most  easily,  of  the  positions  of  objects  from 
viewing  their  perspectives. 

This  we  can  do  with  the  least  difficulty,  when  the  per- 
spective plane  is  taken  parallel  to  the  principal  lines  of  ' 
the  object. 


LIIs’EAR  PERSPECTIVE. 


97 


Of  the  objects  in  nature,  the  larger  portion  of  lines 
are  vertical;  therefore,  in  most  perspective  drawings 
the  perspective  plane  has  a vertical  position. 

64.  Before  an  object  can  be  put  in  perspective  three 
things  must  be  given,  or  known. 

1°.  The  place  of  the  eye. 

2"*.  The  position  of  the  perspective  plane. 

3®.  The  position  of  the  object  to  be  put  in  perspec- 
tive. 

65.  By  considering  what  has  already  been  said,  we 
may  deduce  the  following  principles  : 

T.  If  through  the  eye  and  any  point  in  space,  a visual 
ray  be  drawn,  the  place  at  which  it  pierces  the  perspec- 
tive plane  is  the  perspective  of  the  point. 

2*^.  If  through  the  eye  and  all  the  points  of  a right 
line,  a system  of  visual  rays  be  drawn,  they  will  form  a 
plane  passing  through  the  eye  and  the  right  line ; this 
plane  is  called  a visual  plane,  and  its 'trace  on  the  per- 
spective plane  is  the  perspective  of  the  right  line. 

3®.  If  through  the  eye  and  all  the  points  of  a curve^ 
a system  of  visual  rays  be  drawn,  they  will,  in  general, 
form  the  surface  of  a cone,  the  vertex  being  at  the 
eye ; this  cone  is  called  a visual  cone,  and  its  intersec- 
tion with  the  perspective  plane  is  the  perspective  of  the 
curve. 

66.  In  determining  the  perspective  of  an  object,  it  is 
unnecessary,  and  indeed  impracticable,  to  draw  visual 
rays  through  all  its  points  that  are  seen,  and  to  con- 
struct the  intersections  of  these  rays  with  the  perspective 
plane.  We  therefore  select  the  prominent  points  and 
lines  only,  such  as  the  vertex  and  edges  of  a pyramid, 
the  vertex  of  a cone,  the  edges  of  a prism,  &c. : and 
having  put  these  lines  in  perspective,  we  have,  in  fact, 
determined  the  perspective  of  the  bodv. 

G 


98 


TREATISE  ON 


67.  If  through  the  eye  a system  of  visual  rays  be 
drawn  tangent  to  the  object  to  be  put  in  perspective,  they 
will,  in  general,  form  the  surface  of  a visual  cone  tan- 
gent to  the  object ; the  line  of  contact  is  called  the  ap- 
parent contour  of  the  object ; and  the  intersection  of  the 
surface  of  this  cone  with  the  perspective  plane,  is  the 
boundary  of  the  perspective  of  the  object* 

We  shall  now  apply  these  principles  in  finding  the 
perspectives  of  objects. 

PROBLEM  I. 

Having  given  a cube  and  its  shadow  on  the  horizontal 
plane^  it  is  required  to  find  the  perspective  of  the  cube  and  the 
perspective  of  its  shadow, 

68.  Let  DEGF  (PI.  12)  be  the  horizontal  projection, 
and  D'E  GT'  the  vertical  projection  of  the  cube.  And 
let  J^chgO  be  the  shadow  cast  on  the  horizontal  plane. 

Let  (A,  A')  be  the  place  of  the  eye,  and  BC,  BG  the 
traces  of  the  perspective  plane. 

The  perspective  of  the  cube  and  its  shadow,  after 
they  shall  have  been  found,  will  be  projected  on  the 
planes  of  projection  in  the  traces  of  the  perspective 
plane  (Des.  Geom.  24) ; but  in  order  to  exhibit  their  per- 
spective truly  to  the  eye,  it  must  be  presented  as  it 
appears  on  the  perspective  plane. 

For  this  purpose  we  remove  the  perspective  plane 
parallel  to  itself,  any  convenient  distance,  as  BB',  and 
then  revolve  it  about  its  vertical  trace  B'E'  until  it  coin- 
cides with  the  vertical  plane  of  projection.  The  place 
of  the  eye  is  supposed  to  be  moved  with  the  perspective 
plane,  and  to  have  the  same  relative  position  with  it 
after  it  has  been  revolved. 


LINEAR  PERSPECTIVE. 


99 


Through  the  angle  (D,D')  of  the  cube,  draw  the  visual 
ray  (AD,  AD') ; this  ray  pierces  the  perspective  plane 
at  the  point  (c?,  a'),  and  after  the  plane  has  been  moved 
and  revolved,  the  perspective  of  the  point  is  at  d”. 

Through  (E,  D')  draw  the  visual  ray  (AE,  AD') ; this 
ray  pierces  the  perspective  plane  at  (a,  a),  and  deter- 
mines a"  the  perspective  of  (E,  D').  Hence  d”a  is  the 
perspective  of  DE. 

The  visual  ray  through  the  point  (F,  F^)  pierces  the 
perspective  plane  at  the  point  (/',  /"),  and  determines 
e,  the  perspective  of  the  point  (F,F').  The  visual  ray 
through  the  point  (E,E')  pierces  the  perspective  plane  at 
(«,  ^),  and  determines  k\  the  perspective  of  the  point 
(E,  E').  Hence  ale  is  the  perspective  of  the  edge 
(E,  D'E')  of  the  cube. 

The  visual  ray  through  (D,  E')  pierces  the  perspective 
plane  at  ((/,  ^),  and  determines  the  perspective  of  the 
point  (D,  E').  Hence  pic  is  the  perspective  of  the  edge 
(DE,  E')  of  the  cube,  d''p  of  the  edge  (D,  D'E'),  and 
the  square  alcpd"  of  the  front  face  of  the  cube,  which 
is  parallel  to  the  perspective  plane. 

Having  determined,  by  similar  constructions,  the  per- 
spectives of  the  other  angles  of  the  cube,  we  see  that 
the  trapezoid  d"pqf\^  the  perspective  of  the  face  which 
is  horizontally  projected  in  the  line  DF ; the  trapezoid 
plcnq  the  perspective  of  the  upper  face  of  the  cube ; 
the  trapezoid  d"d'nf  the  perspective  of  the  base  of  the 
cube ; the  square  fnnq  the  perspective  of  the  back  face  of 
the  cube  which  is  parallel  to  the  perspective  plane ; and 
the  trapezoid  ak'nnih^  perspective  of  the-fkce  of  the 
cube  which  is  horizontally  projected  in  the  line  EG. 

There  are  but  three  faces  of  the  cube  which  are  seen, 
viz.  the  upper  face, -the  front  face  parallel  to  the  per- 
spective plane,  and  the  face  which  is  horizontallv  pro- 

G2 


100 


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jected  in  the  line  FD.  The  perspectives  of  the  lines 
bounding  these  faces  are  made  full  in  the  perspective 
plane.  The  perspectives  of  the  other  edges  are  dotted, 
in  order  to  show  how  the  cube  would  appear  if  it  were 
a transparent  body. 

The  perspective  of  the  shadow  on  the  horizontal 
plane  is  found  by  finding  the  perspectives  of  the  points 
c,  and  g*,  and  drawing  the  lines  e h.g  and  gn. 

The  line  hg^  which  is  the  perspective  of  hg^  inter- 
sects  at  c the  line  f q\  which  is  the  perspective  of  the 
edge  (F,  F'G')  of  the  cube.  The  part  eg  of  the  line  hg 
not  seen. 

If  through  the  vertical  edge  (F,  F'G')  of  the  cube,  a 
visual  plane  be  passed,  it  will  cut  the  line  hg  in  the  point 
whose  perspective  is  c.  That  part  of  the  line  hg  lying 
between  the  visual  plane  and  the  vertical  plane  of  pro- 
jection will  not  be  seen,  because  the  cube  intervenes. 

The  visual  ray  passing  through  the  point  whose  per- 
spective is  c,  will  intersect  the  line  hg,  and  the  edge 
(F,  F'G')  of  the  cube  ; and  generally  the  visual  ray  pass- 
ing through  the  point  in  which  the  perspectives  of  two  lines 
intersect,  will  intersect  both  the  lines  in  space, 

69.  This  method  of  perspective  is  often  used  advan- 
tageously in  finding  the  perspectives  of  bodies  bounded 
by  curved  surfaces.  If  through  the  eye  a plane  be 
passed  tangent  to  the  object  to  be  put  in  perspective,  the 
point  of  contact  will  be  a point  of  the  apparent  con- 
tour of  the  object  (67),  and  the  visual  ray  drawn 
through  the  point  of  tangency  will  determine  a point  in 
the  boundary  of  the  perspective.  For  example,  if  it 
were  required  to  find  the  perspective  of  a sphere,  the 
position  of  the  body,  the  perspective  plane,  and  the 
place  of  the  eye  being  given,  we  should  first  pass  through 
the  eye  a system  of  planes  tangent  to  the  sphere,  and 


LINEAR  PERSPECTIVE. 


101 


then  draw  visual  rays  through  the  points  of  contact ; 
the  points  in  Avhich  these  visual  rays  pierce  the  per- 
spective plane  are  points  in  the  boundary  of  the  per- 
spective of  the  sphere. 


CHAPTER  11. 

OF  THE  METHOD  OF  PERSPECTIVE  BY  MEANS  OF  DIAGONALS 
AND  PERPENDICULARS. 

70.  The  point  from  Avhich  the  eye  is  supposed  to  view 
an  object  put  in  perspective,  is  called  the  point  of  sight ; 
and  the  projection  of  this  point  on  the  perspective  plane 
is  called  the  centre  of  the  picture. 

71.  If  through  the  point  of  sight  a right  line  be  drawn 
parallel  to  any  right  line  in  space,  the  point  in  which  it 
pierces  the  perspective  plane  is  called  the  vanishing  point 
of  that  line. 

Hence,  all  parallel  right  lines  have  the  same  vanishing 
point : for  a right  line  parallel  to  one  of  them  will  be 
parallel  to  all  the  others. 

Hence  also,  all  lines  perpendicular  to  the  perspective 
plane  have  their  vanishing  point  at  the  centre  of  the 
picture. 

Regarding  a right  line  as  indefinite  in  length,  its  van 
ishing  point  is  a point  of  its  perspective.  For,  the  par- 
allel through  the  point  of  sight  is  contained  in  the  visual 
plane  passing  through  the  given  line;  therefore,  the  point 
at  which  it  pierces  the  perspective  plane  is  in  the  trace 
of  the  visual  plane.  But  the  trace  of  the  visual  plane 


102 


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is  the  perspective  of’  the  given  line  (65) ; therefore,  the 
vanishing  point  of  a line  is  a point  of  its  perspective ; 
and  is  the  perspective  of  the  point  at  an  infinite  distance 
from  the  perspective  plane. 

When  a line  is  indefinite  in  length,  its  perspective 
is  called  the  indefinite  perspective  of  the  line. 

It  follows  from  what  has  just  been  shown,  that  the 
centre  of  the  picture  is  common  to  the  perspectives  of  all  lines 
which  are  perpendicular  to  the  perspective  plane. 

72.  The  point  in  which  aline  pierces  the  perspective 
plane,  is  also  a point  of  its  perspective.  When  there- 
fore we  have  found  this  point,  and  the  vanishing  point, 
the  perspective  of  the  line  can  be  drawn. 

If  a system  of  lines  be  parallel  to  the  perspective 
plane,  the  line  drawn  through  the  point  of  sight  parallel 
to  them,  will  also  be  parallel  to  the  perspective  plane ; 
and  hence,  the  vanishing  point  of  the  system  will  be  at 
an  infinite  distance  from  the  centre  of  the  picture. 

Now,  since  the  perspectives  of  all  the  lines  must  pass 
through  their  vanishing  point,  and  since  their  vanishing 
point  is  at  an  infinite  distance  from  the  centre  of  the 
picture,  it  follows,  that  the  perspectives  of  lines  so 
situated  are  parallel  to  each  other.  We  can  prove  in 
another  way  that  these  perspectives  are  parallel. 

F or,  the  visual  plane  passing  through  either  of  the 
parallel  lines,  will  intersect  the  perspective  plane  in  a 
line  parallel  to  the  system  of  lines;  hence,  the  traces  of 
the  several  visual  planes  will  be  parallel  to  each  other. 
But  these  traces  are  the  perspectives  of  the  given  lines ; 
therefore,  when  a system  of  parallel  lines  is  parallel  to 
the  perspective  plane,  their  perspectives  will  be  parallel 
to  each  other. 

It  follows,  from  what  has  been  said,  that  when  a line 
is  parallel  to  the  perspective  plane,  the  perspective  of 
the  line  will  be  parallel  to  the  line  itself. 


LINEAR  PERSPECTIVE. 


103 


73.  If  we  have  a system  of  parallel  right  lines,  which 
are  not  parallel  to  the  perspective  plane,  their  perspec- 
tives will  meet  in  a common  point.  F or,  the  system  of 
parallel  lines  will  have  a common  vanishing  point  (71); 
and  the  indefinite  perspective  of  each  line  will  pass 
through  this  point. 

The  same  fact  may  also  be  proved  otherwise.  For, 
suppose  visual  planes  to  be  drawn  through  the  lines. 
These  visual  planes  will  all  intersect  each  other  in  a line 
passing  through  the  eye  and  parallel  to  the  system  of 
lines ; the  point  in  which  this  parallel  pierces  the  per- 
spective plane  will  be  common  to  the  traces  of  all  the 
visual  planes,  and  consequently,  to  the  perspectives  of 
all  the  lines. 

74.  The  horizontal  line  drawn  through  the  centre  of 
the  picture,  is  called  the  horizon  of  the  picture^  and  will 
be  the  locus  of  the  vanishing  points  of  all  horizontal  * 
lines.  F or,  all  horizontal  lines  drawn  through  the  point 
of  sight,  are  contained  in  the  horizontal  plane  through 
the  point  of  sight,  and  will  therefore  pierce  the  per- 
spective plane  in  the  trace  of  this  horizontal  plane, 
which  trace  is  the  horizontal  line  drawn  through  the 
centre  of  the  picture 

75.  Horizontal  lines,  which  make  angles  of  45°  with 
the  perspective  plane,  are  called  diagonals.  If  through 
the  point  of  sight  two  diagonal  lines  be  drawn,  one  on 
each  side  of  the  perpendicular  to  the  perspective  plane, 
the  points  in  which  they  pierce  the  perspective  plane 
are  called  the  vanishing  points  of  diagonals.  These 
points  are  in  the  horizon  of  the  picture,  and  at  equal 
distances  from  the  centre. 

76.  Since  the  line  which  terminates  either  of  these 
points  makes  an  angle  of  45°  with  the  perpendicular 
through  the  point  of  sight,  as  well  as  with  the  perspec- 


104 


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tive  plane,  it  follows  that  the  distance  from  the  centre 
of  the  picture  to  the  point  of  sight  is  equal  to  the  dis- 
tance from  the  centre  of  the  picture  to  the  vanishing 
point  of  diagonals. 

Let  AB  (PI.  13.  Fig.  n)  be  the  trace  of  the  perspec- 
tive plane,  which  is  perpendicular  to  the  plane  of  the 
paper. 

Let  C be  the  horizontal  projection  of  the  point  of  sight, 
and  C'  its  projection  on  the  perspective  plane : C'  is  then 
the  centre  of  the  picture  (70).  Through  the  point  of 
sight  conceive  two  diagonal  lines  to  be  drawn — their 
horizontal  projections  are  Ca  and  C^,  which  make  angles 
of  45°  with  the  perpendicular  CD.  But  these  diagonals 
pierce  the  perspective  plane  at  the  points  d and  in  the 
horizontal  line  dCb\  and  these  points,  by  definition,  are 
the  vanishing  points  of  diagonals.  Since  the  angles 
DCa  and  DC5  are  each  45°,  Da  and  D6  are  each  equ-al  to 
CD.  But  Cd  and  CU  are  each  equal  to  Da  or  D^,  and 
consequently  to  DC,  which  is  the  distance  of  the  point 
of  sight  from  the  centre  of  the  picture. 

77.  By  considering  what  has  been  said,  we  have  this 
general  rule  for  determining  whether  the  vanishing 
point  of  a diagonal  is  on  the  right  or  left  of  the 
centre  of  the  picture.  Through  any  point  of  the  dia- 
gonal, in  front  of  the  perspective  plane,  draw  a perpen- 
dicular to  the  perspective  plane.  Now,  when  the  part 
of  the  diagonal  intercepted  between  the  point  and  the 
perspective  plane,  lies  on  the  right  of  the  perpendicular, 
the  vanishing  point  of  the  diagonal  is  on  the  right  of  the 
centre  of  the  picture ; but  when  it  lies  on  the  left  of  the 
perpendicular,  the  vanishing  point  is  on  the  left  of  the 
centre  of  the  picture.  The  rule  is  reversed  when  the 
point  through  which  the  perpendicular  is  drawn  is  behind 
the  perspective  plane. 


LINEAR  PERSPECTIVE. 


lot 

78.  The  principles  on  which  the  perspectives  of  points 
are  found,  by  the  method  we  are  now  explaining,  are 
these, 

r.  That  the  perspective  of  every  point  of  a right 
line  is  found  somewhere  in  the  indefinite  perspective  of 
the  line. 

2"*.  That  the  perspectives  of  two  points  of  a right 
line  determine  the  perspective  of  the  line.  From  the 
first  principle  it  follows,  that  if  two  lines  be  drawn 
through  any  point  in  space,  and  their  perspectives  de- 
termined, the  intersection  of  their  perspectives  will  be 
the  perspective  of  their  point  of  intersection. 

Let  c (PI.  13,  Fig.  7i)  be  a point  in  the  horizontal 
plane  at  a distance  c d behind  the  perspective  plane. 

The  perpendicular  through  c pierces  the  perspective 
plane  ate/;  and  since  C'  is  the  vanishing  point  of  perpen- 
diculars, dC  is  the  perspective  of  the  perpendicular  cd. 
The  diagonal  through  c pierces  the  perspective  plane  at 
/,  and  the  diagonal  has  its  vanishing  point  at  a , there- 
fore fd  is  its  perspective.  But  the  perspective  of  the 
point  c is  found  both  in  dO  and  mfd  ; it  is  therefore  at 
c their  point  of  intersection. 

We  could  have  determined  the  perspective  of  the 
point  c by  finding  the  perspectives  of  any  other  two  lines 
passing  through  it ; but  it  is  better  to  use  the  perpen- 
dicular and  diagonal  than  other  lines,  because  their 
perspectives  are  more  readily  found. 

79.  The  perspective  plane  being  placed  between  the 
eye  and  the  object,  the  perspective  of  the  object  will 
lie  above  the  ground  line,  and  its  horizontal  projection 
will  be  behind  the  perspective  plane.  Now,  when  the  hori- 
zontal plane  is  revolved  in  the  usual  way,  to  coincide 
with  the  perspective  plane,  the  perspective,  the  horizon- 


106 


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tal  projection,  and  the  vertical  projection  of  the  object 
will  occupy  the  same  part  of  the  paper. 

To  avoid  this  inconvenience  in  the  constructions,  we 
revolve  the  horizontal  plane  of  projection  about  its  inter- 
section with  the  perspective  plane,  in  such  a manner 
that  the  part  behind  the  perspective  plane  shall  fall 
below  the  ground  line.  When  this  is  done,  the  diagon- 
als will  have  a direction  in  the  construction  contrary  to 
their  true  direction  in  space. 

F or  example,  had  we  so  revolved  the  horizontal  plane 
before  finding  the  perspective  of  the  point  c,  the  horizon- 
tal projection  of  instead  of  being  above  the  ground 
line  at  c,  would  have  been  below  it  at  dg  being  equal 
to  dc.  In  this  case  the  diagonal  would  intersect  the 
ground  line  at  the  point  /,  as  before,  and  the  perpendi- 
cular would  meet  the  ground  line  at  d.  If  then,  we 
draw  dC  and  fa\  we  obtain  c\  the  perspective  of  the 
point  as  before.  But  there  is  this  difterence  ; were  we 
to  consider  the  point  g in  front  of  the  perspective  plane, 
which  we  ought  to  do  if  the  horizontal  plane  were  re- 
volved in  the  usual  way,  the  diagonal  g*/ would  hav.e  its 
vanishing  point  at  b\  and  the  perspective  of  the  point 
would  be  below  the  ground  line. 

We  may  then  project  points  which  are  behind  the  per- 
spective plane,  on  the  part  of  the  paper  in  front  of  the 
ground  line,  by  observing  that  the  true  vanishing  point 
of  any  diagonal  is  the  one  opposite  to  that  which  its 
projection  indicates. 

80.  If  through  the  point  of  sight  a plane  be  passed 
parallel  to  any  plane  in  space,  it  will  contain  all  lines 
drawn  through  the  point  of  sight  and  parallel  to  the  lat- 
ter plane.  Hence,  the  trace  of  the  visual  plane  will  be 
the  locus  of  the  vanishing  points  of  all  lines  contained 
in  the  parallel  plane. 


LINEAR  PERSPECTIVE. 


107 


Therefore,  the  vanishing  line  of  any  'plane  is  the  trace; 
on  the  perspective  plane,  of  the  visual  plane  drawn 
parallel  to  it. 


PROBLEM  II. 

Having  four  cubes  placed  in  the  four  angles  of  a square^ 
their  bases  in  the  same  horizontal  plane^  it  is  required  to  find 
the  perspective  of  the  cubes  and  the  perspective  of  their  shad- 
ows on  the  plane  of  their  bases, 

81.  If  the  perspective  plane  were  taken  through  the 
front  faces  of  the  cubes  whose  horizontal  projections 
are  cr  and  ds  (PI.  13,  Fig.  m),  AC  would  be  its  horizon- 
tal trace,  and  the  front  faces  of  the  cubes  being  in  the 
perspective  plane,  would  be  their  own  perspectives. 
The  projection  in  this  figure,  although  made  on  a small 
scale,  shows  the  position  which  the  cubes  have  with 
each  other,  and  with  the  perspective  plane. 

Let  us  also  suppose  the  point  of  sight  to  be  in  a plane 
equidistant  from  the  inner  faces  of  the  cubes. 

Draw  any  line,  as  CD,  for  the  ground  line  of  the  plane 
on  which  the  perspective  is  to  be  made. 

Assume  E for  the  centre  of  the  picture,  draw  the  ho- 
rizontal line  H'EH,  and  take  H and  H'  for  the  vanishing 
points  of  diagonals.  Then  EH,  or  EH'  is  equal  to  the 
distance  of  the  point  of  sight  from  the  perspective 
plane. 

Draw  EE'  perpendicular  to  the  ground  line,  and 
from  E'  lay  off*  E'^  and  E'c,  each  equal  to  half  the  dis 
tance  between  the  inner  faces  of  the  cubes.  Make  db 
and  ca  each  equal  to  the  length  of  an  edge  of  the  given 
cubes,  and  on  them  construct  the  squares  d U and  c d ; 
these  squares  are  the  perspectives  of  the  faces  which 


108 


TREATISE  ON 


are  in  the  perspective  plane.  Through  the  points 
</,  5,  a 5 e , d'  and  b\  draw  lines  to  E,  the  centre  of  the  pic- 
ture. These  lines  are  the  indefinite  perspectives  of  the 
lines  aq^  cp^dl^  bh  (Fig.  m),  and  of  the  parallel  edges 
directly  over  them  (72). 

Through  a and  b draw  the  lines  aH  and  to  the 
vanishing  points  of  diagonals  ; these  lines  are  the  indefi- 
nite perspectives  ah  and  b Fig.  m. 

The  points  e,  m,/  andg*,  in  which  the  perspectives  of 
the  diagonals  intersect  the  perspectives  of  the  perpen- 
diculars, are,  respectively,  the  perspective  of  the  points 
e,  m,  /,  and  Fig.  m. 

Through  e draw  c c parallel  to  cc;  the  point  e,  in 
which  it  intersects  c'E,  is  the  perspective  of  the.  angular 
point  of  the  cube  which  is  horizontally  projected  at  c. 
Fig.  m.  F or  the  edge  of  the  cube  which  pierces  the  hori- 
zontal plane  at  e is  parallel  to  the  edge  which  pierces  it 
at  6',  and  both  are  parallel  to  the  perspective  plane; 
hence  their  perspectives  are  parallel  (73).  But  both 
these  edges  are  limited  by  the  edge  which  pierces  the 
perspective  plane  at  c\  and  the  indefinite  perspective  of 
this  latter  edge  is  c E ; therefore  e is  the  perspective  of 
the  angular  point  directly  over  the  one  whose  perspective 
is  e.  Similar  reasoning  will  apply  to  the  other  vertical 
edges  of  the  cubes. 

Through  e draw  e r parallel  to  a c ; the  point  r,  where  it 
meets  aE  is  the  perspective  of  the  angular  point  r.  Fig. 
m.  From  r draw  r r parallel  to  a a,  and  from  the  point 
r where  it  intersects  a'E  draw  re. 

We  have  then  determined  the  square  accd,^  the  per- 
spective of  the  face  in  the  perspective  plane  : the  trape- 
zoid a c e r,  the  perspective  of  the  base  of  the  cube : the 
trapezoid  c e e c,  the  perspective  of  a face  perpendicular 
to  the  perspective  plane : the  square  reer\  the  perspec- 


LINEAR  PERSPECTIVE. 


109 


jve  of  the  face  parallel  to  the  perspective  plane ; the 
trapezoid  cerd^  the  perspective  of  the  upper  face  of  the 
cube:  and  the  trapezoid  arrd^  the  perspective  of  a 
second  face  perpendicular  to  the  perspective  plane. 

The  perspective  of  the  angular  point  m of  the  cube  mq^ 
Fig.  m,  has  already  been  determined ; and  since  the  edge 
mm^  Fig.  m,  is  parallel  to  the  perspective  plane,  the  per- 
spective of  the  point  n must  lie  in  m w,  drawn  through 
parallel  to  the  ground  line  but  it  is  also  in  aE,  the  in- 
definite perspective  of  aq;  hence  it  is  at  w,  their  point 
of  intersection. 

Through  m and  n draw  vertical  lines  and  produce 
them  till  they  meet  the  lines  e E and  a E,  and  join  the  points 
of  intersection;  we  have  then  the  perspective  of  the 
face  parallel  to  the  perspective  plane. 

Through  draw  nH  to  the  vanishing  point  of  diagonals ; 
this  line  is  the  perspective  of  np^  Fig.  m,  and  the  point  p, 
where  it  intersects  cE,  is  the  perspective  of  the  point  jo, 
Fig.  m.  Through  p draw  pq  parallel  to  the  ground  line 
till  it  meets  aE;  and  at  q and^  erect  perpendiculars  to 
the  ground  line,  and  complete  the  perspective  of  the 
cube  as  in  the  last  case.  The  perspectives  of  the  other 
cubes  are  determined  in  a manner  entirely  similar. 

It  remains  to  find  the  perspective  of  the  shadows  cast 
on  the  horizontal  plane. 

82.  The  shadow  which  a point  casts  upon  a plane,  is 
always  found  in  the  ray  of  light  passing  through  the 
point,  and  also  in  the  projection  of  this  ray  upon  the 
plane  on  which  the  shadow  falls.  Hence,  the  perspec- 
tive of  the  shadow  will  be  found  in  the  perspective  of  the 
ray,  and  in  the  perspective  of  its  projection,  and  is  con- 
sequently their  point  of  intersection. 

83.  The  perspective  of  the  shadow  cast  by  a right  line 


ilO 


TREATISE  ON 


on  a plane,  is  in  the  indefinite  perspective  of  the  inter 
section  of  a plane  of  rays  passed  through  the  line,  with 
the  plane  on  which  the  shadow  falls. 

84.  Since  the  rays  of  light  are  parallel,Jthey  have  a 
common  vanishing  point  (71).  This  point  is  where  a 
ray  of  light  drawn  through  the  point  of  sight  pierces  the 
perspective  plane. 

Let  (E,  E'),  Fig.  be  the  direction  of  a ray  of  light. 
Through  the  point  of  sight  (C,  C')  let  a ray  of  light  be 
drawn,  it  will  pierce  the  perspective  plane  at  R,  which  is, 
therefore,  the  vanishing  point  of  rays. 

The  projections  of  rays  on  any  plane  are  also  parallel 
to  each  other,  and  consequently  they  have  a common 
vanishing  point. 

Through  the  point  of  sight  let  a line  be  drawn  parallel 
to  E,  the  horizontal  projection  of  a ray  of  light.  This 
parallel  pierces  the  perspective  plane  at  F,  which  is, 
therefore,  the  vanishing  point  of  the  horizontal  projec- 
tions of  rays. 

Now,  since  the  vertical  plane  of  rays  through  the 
point  of  sight  contains  the  ray  of  light  through  the 
point  of  sight,  and  also  the  line  drawn  parallel  to  the 
horizontal  projections  of  rays,  it  follows,  that  its  trace 
on  the  perspective  plane  will  contain  both  the  vanishing 
point  of  rays  and  the  vanishing  point  of  horizontal  pro- 
jections. But  its  trace  on  the  perspective  plane  is  a 
vertical  line ; hence,  the  line  joining  the  vanishing  point  of 
rays  and  the  vanishing  point  of  horizontal  projections^  is  per- 
pendicular to  the  ground  line, 

85.  We  see  that  the  vanishing  point  of  rays  and  the 
vanishing  point  of  horizontal  projections  can  be  found 
when  we  know  the  direction  of  the  light.  Reciprocally, 
if  we  know  the  vanishing  point  of  rays  and  the  vanish- 


LINEAR  PERSPECTIVE. 


Ill 


ing  point  of  horizontal  projections,  we  can  determine 
the  direction  of  the  light. 

For,  the  line  joining  the  vanishing  point  of  rays  and 
the  centre- of  the  picture,  will  be  parallel  to  the  projec- 
tions of  the  rays  on  the  perspective  plane.  And  if 
through  the  centre  of  the  picture,  the  perpendicular  C'D 
be  drawn  to  the  ground  line,  and  produced,  and  DC 
made  equal  to  the  distance  from  the  centre  of  the  pic- 
ture to  the  vanishing  point  of  diagonals,  we  shall  have 
C the  projection  of  the  point  of  sight  on  the  horizontal 
plane.  Joining  this  point  with  F,  the  point  in  which  the 
line  joining  the  vanishing  point  of  rays  and  the  vanishing 
point  of  horizontal  projections  intersects  the  ground  line, 
and  we  have  CF'  the  horizontal  projection  of  the  ray  of 
light  passing  through  the  point  of  sight. 

86.  In  finding  the  shadows  cast  by  the  cubes  on  the 
horizontal  plane,  let  R be  the  vanishing  point  of  rays, 
and  P the  vanishing  point  of  horizontal  projections  of 
rays. 

By  considering  the  direction  of  the  light,  it  is  plain, 
that  the  edge  of  the  cube  whose  perspective  is  c c will 
cast  a shadow  on  the  horizontal  plane.  But  the  plane 
of  rays  through  this  edge  is  perpendicular  to  the  hori- 
zontal plane ; therefore,  the  shadow  of  the  edge  is  parallel 
to  the  horizontal  projection  of  the  rays,  and  consequently 
P is  its  vanishing  point.  But  the  shadow  is  limited  by 
the  ray  through  the  point  whose  perspective  is  c;  there- 
fore, its  perspective  is  limited  by  the  perspective  of  this 
ray,  which  is  c'R : consequently,  c t is  the  perspective  of 
the  shadow. 

The  next  line  which  casts  a shadow  on  the  horizontal 
plane  is  the  edge  whose  perspective  is  ce.  Since  this 
line  is  perpendicular  to  the  perspective  plane,  it  will  be 
parallel  to  the  horizontal  plane,  and  therefore  its  shadow, 


112 


TREATISE  ON 


which  1$  parallel  to  itself,  will  be  perpendicular  to  the 
perspective  plane ; and  hence,  its  vanishing  point  is  at 
E.  But  t is  one  point  of  its  perspective ; therefore  is 
its  indefinite  perspective.  This  perspective  is  limited 
by  e'R  drawn  to  the  vanishing  point  of  rays. 

The  shadow  cast  by  the  edge  whose  perspective  is 
re  is  parallel  to  the  line  itself,  and  also  to  the  perspec- 
tive plane ; hence  its  indefinite  perspective  is  n drawn 
through  n parallel  to  the  ground  line.  This  perspective 
is  limited  by  r R,  and  also  by  rP.  The  line  rz;  is  the  per- 
spective of  the  shadow  cast  by  the  vertical  edge  of  the 
cube  whose  perspective  is  r r . Only  a small  part  of  this 
shadow  is  seen,  nearly  all  of  it  being  behind  the  cube. 

The  shadows  of  the  other  cubes  are  found  in  a manner 
so  entirely  similar,  as  not  to  require  particular  expla- 
nations. 

The  faces  of  the  cubes  which  are  in  the  shade  are 
darkened  in  the  perspective. 


PROBLEM  III. 

Having  given  four  pyramids^  standing  on  pedestals^  and 
situated  in  the  four  angles  of  a square^  it  is  required  to  find 
the  perspectives  of  the  pyramids  and  pedestals^  and  the  per- 
spectives of  their  shadows. 

87.  Fig.  n (PL  1 4)  represents,  on  a small  scale,  the 
projections  of  the  four  pyramids  and  pedestals,  having 
the  same  relative  position  as  those  which  are  to  be  put 
in  perspective. 

Let  the  perspective  plane  be  taken  through  the  front 
faces  of  the  pedestals : AB  is  its  horizontal  trace. 

From  any  point,  as  a,  lay  off  a 6 equal  to  the  side  of 
the  square  which  forms  the  base  of  the  pedestal,  and  on 


LINEAR  PERSPECTIVE. 


113 


it  describe  a rectangle  equal  to  the  front  face  of  the 
pedestal.  From  b lay  off  6 c equal  to  the  distance  be- 
tween the  pedestals,  and  make  cd  equal  to  ab;  and  on  cc? 
describe  a rectangle  equal  to  the  front  faces  of  the 
pedestals.  The  perspectives  of  the  four  pedestals  are 
then  determined  by  constructions  entirely  similar  to  those 
of  the  last  problem. 

To  find  the  perspectives  of  the  vertices  of  the  pyra- 
mids : 

Let  S be  the  centre  of  the  picture,  and  D and  D the 
vanishing  points  of  diagonals. 

If  a line  be  drawn  through  the  vertices  of  the  two 
pyramids  on  the  left,  it  will  be  perpendicular  to  the  per- 
spective plane,  and  will  pierce  it  in  the  vertical  line  h h\ 
drawn  through  the  middle  point  of  a b.  On  this  line^ 
therefore,  lay  off  h K equal  to  the  height  of  the  vertices 
of  the  pyramids  above  the  base  of  the  pedestals,  and  K 
will  be  the  point  in  which  the  line  through  the  vertices 
of  the  pyramids  pierces  the  perspective  plane.  The 
line  HS  is  the  perspective  of  this  perpendicular. 

The  diagonal  through  the  vertex  of  the  front  pyramid 
pierces  the  perspective  plane  in  the  horizontal  line  1ia\ 
and  also  in  the  vertical  line  ad ; hence  a D'  is  the  per- 
spective of  the  diagonal,  and  v the  perspective  of  the 
""ertex  of  the  front  pyramid.  The  perspectives  of  the 
vertices  of  the  other  pyramids  are  easily  found. 

From  e lay  off  ef  equal  to  the  distance  which  the 
pedestal  projects  beyond  the  pyramid,  draw  e\}'  to  the 
vanishing  point  of  diagonals,  and  from  / draw /S  to  the 
vanishing  point  of  perpendiculars ; the  point  in  which 
these  lines  intersect,  is  the  perspective  of  the  point  in 
which  an  edge  of  the  pyramid  pierces  the  upper  face  of 
the  pedestal,  and  the  horizontal  line  gp  is  the  indefinite 

H 


114 


TREATISE  ON 


perspective  of  the  line  in  which  the  front  face  of  the 
pyramid  intersects  the  upper  face  of  the  pedestal. 

From  i lay  off  on  z e a distance  equal  to  e and 
through  the  point  so  determined  draw  a line  to  the  centre 
of  the  picture : the  point  p,  in  which  it  meets  gjo,  is  the 
perspective  of  the  point  in  which  a second  edge  of  the 
pyramid  pierces  the  upper  face  of  the  pedestal.  The 
diagonals  through  p and  g determine,  by  their  intersec- 
tions with  the  perpendiculars  g-S  andpS,  the  points  in 
which  the  two  remaining  edges  pierce  the  upper  face  of 
the  pedestal.  Joining  these  points  with  v,  the  perspec- 
tive of  the  vertex,  we  have  the  perspective  of  the  pyramid. 

Only  a part  of  the  edge  of  the  pedestal  which  is  per- 
pendicular to  the  perspective  plane  at  c,  is  seen.  The 
perspective  of  the  edge  intersects  the  perspective  of  the 
edge  of  the  pyramid  at  L If  through  the  point  I a 
visual  ray  be  drawn,  it  will  intersect,  in  space,  both  the 
edge  of  the  pyramid  and  the  edge  of  the  pedestal.  At 
the  point  in  which  it  intersects  the  edge  of  the  pedestal, 
the  edge  of  the  pedestal  passes  behind  the  pyramid,  and 
is  not  seen  •,  arid  the  same  may  be  said  of  the  edge  of 
the  pedestal  parallel  to  e i.  The  perspectives  of  the 
other  pedestals  a-nd  pyramids  are  found  by  constructions 
entirely  similar. 

To  find  the  perspectives  of  the  shadows  cast  on  the 
horizontal  plane : 

Let  R be  the  vanishing  point  of  rays,  and  P the 
vanishing  point  of  horizontal  projections. 

The  point  v',  in  which  the  diagonals  pY)  and 
intersect,  is  the  perspective  of  the  projection  of  the 
vertex  of  the  pyramid  on  the  plane  of  the  upper  base  of 
the  pedestal ; hence,  vT  is  the  perspective  of  the  pro 
jection  of  the  ray  through  the  vertex  of  the  pyramid,  on 
that  plane.  But  vR  is  the  perspective  of  the  ray; 


LINEAR  PERSPECTIVE. 


115 


nence  t"  is  the  perspective  of  the  shadow  cast  by  the 
vertex  of  the  pyramid  on  the  plane  of  the  upper  face 
of  the  pedestal. 

It  is  plain  that  the  edges  of  the  pyramid  which  pierce 
the  upper  face  of  the  pedestal  in  the  points  whose  per- 
spective are  p and  k,  will  cast  shadows  on  the  pedestal 
and  on  the  horizontal  plane.  Therefore, ^pv"  and  kv"  are 
the  indefinite  shadows  on  the  upper  face  of  the  pedestal. 
It  is  evident,  that  only  the  parts  pm  and  kn  fall  on  the 
pedestal,  and  that  the  points  m and  n cast  shadows  on 
the  horizontal  plane.  , 

The  point  r,  in  which  the  diagonals  aD ' and  hD  inter- 
sect, is  the  perspective  of  the  projection  of  the  vertex  of 
the  pyramid  on  the  plane  of  the  base  of  the  pedestal. 
From  r,  draw  rP  to  the  vanishing  point  of  projections, 
and  the  point  v",  in  which  it  intersects  vR,  is  the  per- 
spective of  the  shadow  cast  by  the  vertex  of  the  pyramid 
on  the  horizontal  plane. 

The  line  bi'  is  the  perspective  of  the  shadow  cast  on 
the  horizontal  plane  by  the  edge  b i of  the  pedestal,  and 
2'S  is  the  indefinite  perspective  of  the  shadow  cast  by  is. 
Therefore,  drawing  through  m,  mR  to  the  vanishing  point 
of  rays,  determines  m',  the  shadow  cast  on  the  horizontal 
plane  by  the  point  m:  the  shadow  tm'  is  cast  by  ^ m. 

The  part  m^,  of  the  edge  w,  will  not  cast  a shadow  on 
the  horizontal  plane,  being  itself  in  the  shadow  of  the 
pyramid.  If,  however,  we  draw  from  s a line  to  the 
vanishing  point  of  rays,  the  point  in  which  it  intersects 
^'S,  limits  the  shadow  which  would  fall  on  the  horizontal 
plane  if  the  edge  were  in  the  light.  The  line  drawn 
through  /,  the  point  so  determined,  and  parallel  to  the 
ground  line,  is  the  indefinite  perspective  of  the  shadow 
cast  by  the  edge  X^.  Through  n draw  nR,  and  we  de- 
termine w , the  shadow  cast  on  the  horizontal  plane  by 

H 2 


116 


TREATISE  ON 


the  point  n ; and  un'"  is  the  shadow  cast  by  the  edge  of 
the  pyramid.  The  part  ns,  of  the  edge  of  the  pedestal, 
being  in  the  shadow  of  the  pyramid,  cannot  cast  a 
shadow  on  the  horizontal  plane ; and  the  perspective  of 
the  shadow  cast  by  begins  at  n,  and  is  parallel  to  the 
ground  line. 

The  perspectives  of  the  shadows  of  the  other  pyra- 
mids are  found  by  similar  constructions.  The  faces  of 
the  pyramids  and  pedestals  which  are  in  the  shade  and 
seen,  are  shaded  in  the  drawing. 

PROPOSITION  IV.  THEOREM. 

If  a right  line  be  tangent  to  a curve  in  space,  the  perspective 
of  the  right  line  will  be  tangent  to  the  perspective  of  the  curve. 

88.  For  let  AFCG  (PI.  15,  Fig.  1)  be  a curve,  to  which 
a right  line  is  drawn  tangent  at  any  point  as  F. 

Suppose  a visual  cone  to  be  drawn  through  the  curve, 
and  a visual  plane  through  the  right  line,  the  visual 
plane  will  be  tangent  to  the  visual  cone.  The  perspec- 
tive plane  will  intersect  the  visual  plane  in  a right  line 
and  the  visual  cone  in  a curve,  and  the  right  line  and 
curve  will  be  tangent  to  each  other  (Des.  Geom.  84). 
But  the  right  line  in  which  the  perspective  plane  inter- 
sects the  visual  plane  is  the  perspective  of  the  tangent 
line,  and  the  curve  in  which  the  perspective  plane  inter- 
sects the  visual  cone  is  the  perspective  of  the  given  curve 
AFCG : hence,  when  a right  line  and  curve  are  tangent 
in  space,  their  perspectives  are  also  tangent. 

If  two  curves  are  tangent  to  each  other  in  space,  their 
perspectives  are  also  tangent.  For  the  two  visual 
cones  which  determine  their  perspectives  are  tangent 
to  each  other,  and  therefore  the  curves  in  which  they 
intersect  the  perspective  plane  are  likewise  tangent. 


LINEAR  PERSPECTIVE. 


117 


PROBLEM  V. 

find  the  perspective  of  a circle, 

89.  Let  B (PI.  15,  Fig.  1),  be  the  centre  of  the  circle 
which  is  to  be  put  in  perspective,  and  RT  the  trace  of 
the  perspective  plane ; the  perspective  plane  being  per* 
pendicular  to  the  plane  of  the  circle  AFCG. 

Although  the  horizontal  projection  of  the  circle  is 
made  in  front  of  the  perspective  plane,  all  the  points  of 
it  are,  in  fact,  as  far  behind  it  as  they  are  now  projected 
in  front  of  the  trace  RT. 

Let  the  point  of  sight  be  taken  in  a plane  passing 
through  the  centre  of  the  circle,  and  perpendicular  to 
the  perspective  and  horizontal  planes. 

Let  S be  the  centre  of  the  picture,  and  P and  F 
the  vanishing  points  of  diagonals.  Through  B draw 
the  diagonal  BN,  and  also  a perpendicular  to  the 
ground  line. 

If  the  circle  were  in  front  of  the  perspective  plane, 
the  diagonal  BN  would  have  its  vanishing  point  at  P ; 
but  since  it  is  behind  it,  the  vanishing  point  of  the 
diagonal  is  at  F (79).  Hence  NP'  is  the  perspective 
of  the  diagonal.  The  perpendicular  through  B pierces 
the  perspective  plane  at  w,  and  has  its  vanishing  point 
at  S ; therefore  the  point  h where  Sn  intersects  the  per- 
spective of  the  diagonal,  is  the  perspective  of  the 
centre  B. 

Through  6 draw  the  horizontal  line  ahc]  this  line 
is  the  indefinite  perspective  of  the  diameter  ABC. 
Through  A and  C draw  tangent  lines.  These  tangents 
are  perpendicular  to  the  ground  line,  and  their  perspec- 
tives pass  through  S.  The  points  a and  c in  which 


118 


TREATISE  ON 


they  intersect  the  perspective  of  AC,  are  the  perspectives 
of  the  points  A and  C.  The  lines  Sa  and  Sc  are  tan- 
gent to  the  ellipse,  which  is  the  perspective  of  the  circle 
AFCG. 

Let  the  perspectives  of  the  points  F and  G be  next 
found ; they  are  / and  g. 

If  through  the  points  F and  G tangent  lines  be  drawn 
to  the  circle  AFCG,  their  perspectives  will  be  tangent 
to  the  perspective  of  the  circle  (88).  But  since  the 
tangents  are  parallel  to  the  ground  line,  their  perspec- 
tives will  also  be  parallel  to  the  ground  line ; hence  g/ 
is  perpendicular  to  the  tangents  drawn  through  its  ex- 
tremities/ and^;  it  is,  therefore,  an  axis  of  the  ellipse, 
which  is  the  perspective  of  the  circle. 

Bisect  gf  at  d.  Through  d draw  Fd  e,  and  from  e draw 
the  diagonal  eD.  It  is  plain  that  Fd  e is  the  perspective 
of  the  diagonal  Dc,  and  that  d is  the  perspective  of  D. 

Through  D draw  KDH  parallel  to  the  ground  line, 
and  find  the  perspectives  of  the  points  K and  H,  which 
are  k and  h;  kh\^  the  perspective  of  KH,  and  is  the 
other  axis  of  the  ellipse.  The  ellipse  therefore  can  be 
described. 

90.  When  the  point  of  sight  is  not  in  the  plane  pass- 
ing through  the  centre  of  the  given  circle  and  perpen- 
dicular to  the  perspective  and  horizontal  planes,  we  are 
unable  to  find  the  axes  of  the  ellipse  by  a direct  construc- 
tion. We  then  find  the  perspectives  of  several  points 
of  the  circumference  of  the  circle,  and  describe  the 
ellipse  through  them. 

In  PI.  15,  Fig.  2,  the  perspective  of  the  circle  is  found 
by  points.  The  perspectives  of  the  tangents  at  the 
points  a,  d^  6,  g,  c and  e are  tangent  to  the  perspective 
of  the  circle  at  the  points  a , d\  g\  c and  e. 

91.  Having  given  a circle  in  space,  and  the  point  of 


LINEAR  PERSPECTIVE. 


119 


sight,  we  may  so  place  the  perspective  plane  that  the 
perspective  of  the  circle  shall  be  any  one  of  the  conic 
sections. 

F or,  when  the  circle  and  point  of  sight  are  given,  the 
visual  cone  circumscribing  the  circle  is  also  given,  and 
if  the  position  of  the  perspective  plane  be  undetermined, 
it  may  be  so  chosen  as  to  intersect  the  cone  in  any 
one  of  the  conic  sections. 

When  the  perspective  plane  is  parallel  to  the  base  of 
the  visual  cone,  or  when  it  cuts  the  cone  in' a sub-con- 
trary section,  the  curve  of  intersection  is  a circle. 


PROBLEM  VI. 

To  find  the  perspective  of  a cylinder^  the  perspective  of  the 
shadow  cast  by  the  upper  circle  on  the  interior  surface^  and 
the  perspective  oj  the  shadow  on  the  horizontal  plane, 

92.  Let  the  circle  described  in  the  horizontal  plane 
with  the  centre  C (PI.  15,  Fig.  3),  and  radius  CB,  be  the 
lower  base  of  the  cylinder ; the  centre  C being  at  a dis- 
tance behind  the  perspective  plane  equal  to  CC".  Letx4'B' 
be  the  projection  on  the  perspective  plane  of  the  upper 
base  of  the  cylinder,  the  plane  of  this  base  intersecting 
the  perspective  plane  in  the  horizontal  line  A'B'. 

Let  the  point  of  sight  be  taken  in  the  plane  through 
the  axis  of  the  cylinder  and  perpendicular  to  the  per- 
spective plane.  Let  S be  the  centre  of  the  picture,  and 
P'  and  P the  vanishing  points  of  diagonals. 

Find  now  the  perspective  of  the  lower  base  of  the 
cylinder  as  in  Prob.  5. 

In  finding  the  perspective  of  the  upper  base  we  have 
merely  to  regard  the  perpendiculars  and  diagonals 
already  drawn,  as  the  projections  on  the  horizontal  plane 


J20 


TREATISE  ON 


of  corresponding  perpendiculars  and  diagonals  drawn  in 
the  upper  base  of  the  cylinder. 

For  example,  the  diagonal  being  considered  in 
the  upper  base  of  the  cylinder,  would  pierce  the  perspec- 
tive plane  at  b\  its  perspective  would  be  UaV\  and  its 
intersection  with  C'S  determines  d\  a point  in  the  per- 
spective of  the  upper  base;  the  diagonal  Be  determines 
the  point  c". 

If  through  the  point  of  sight  two  tangent  planes  were 
drawn  to  the  cylinder,  they  would  touch  it  in  the  two 
elements  which  pierce  the  horizontal  plane  at  / and  g. 

Having  found  g and f\  the  perspectives  of  the  points 
g and  f,  in  the  lower  base  of  the  cylinder,  and  g"  and  f'* 
the  perspectives  of  the  corresponding  points  of  the  upper 
base,  draw  the  lines  gg"  and  f'f" ; these  lines  are  the  per- 
spectives of  the  elements  which  pierce  the  horizontal 
plane  at  g and  f. 

The  part  of  the  cylinder  convex  ‘towards  the  point 
of  sight,  and  limited  by  these  elements,  is  seen ; the 
other  part  is  not  seen.  Therefore,  the  semi-ellipses  gef' 
and  gcf\  which  are  seen,  are  made  full,  and  the  semi- 
ellipses gdf  and  gdf'\  which  are  not  seen,  are  dotted. 

To  find  the  perspective  of  the  shadow  cast  on  the 
interior  of  the  cylinder  by  the  circumference  of  the 
upper  base : 

Let  R be  the  vanishing  point  of  rays,  and  H the 
vanishing  point  of  horizontal  projections. 

If  two  tangent  planes  of  rays  be  drawn  to  the  cylin- 
der in  space,  their  horizontal  traces  will  be  tangent  to 
the  base  of  the  cylinder,  and  the  elements  of  contact 
will  be  the  elements  of  shade.  But  the  horizontal  traces 
of  these  planes  will  be  parallel  to  the  horizontal  projec- 
tion of  the  rays  of  light ; hence,  their  vanishing  point 
is  at  H,  The  horizontal  traces  are  also  tangent  to  the 


LINEAR  PERSPECTIVE.  121 

base  of  the  cylinder,  therefore  their  perspectives  will  be 
tangent  to  the  perspective  of  the  base  (B8). 

Through  H draw  the  tangents  Wc  and  H/i ; the  points 
of  contact  k and  h are  the  perspectives  of  the  two 
points  in  which  the  elements  of  shade  pierce  the  hori- 
zontal plane.  But  since  the  elements  of  shade  are  ver- 
tical lines,  kk'  and  hh\  drawn  perpendicular  to  the 
ground  line,  are  their  perspectives,  and  k'  and  K are  the 
perspectives  of  the  points  at  which  the  shadow  on  the 
interior  of  the  cylinder  begins.  ' 

If  we  suppose  the  cylinder  in  space  to  be  intersected 
by  a plane  of  rays  parallel  to  its  axis,  the  horizontal 
trace  of  the  plane  will  be  parallel  to  the  horizontal  pro- 
jection of  the  rays  of  light,  and  consequently,  will  have 
its  vanishing  point  at  H.  Every  plane  so  drawn  will  in- 
tersect the  cylinder  in  two  elements,  and  the  one  towards 
the  source  of  light  will  cast  a shadow  on  the  other. 

Through  H draw  any  line,  as  Hn  2,  to  represent  the 
perspective  of  the  horizontal  trace  of  a secant  plane  of 
rays.  Through  the  points  i and  in  which  it  intersects 
the  perspective  of  the  base,  draw  the  elements  i i and 
n n.  Through  i'  draw  t R to  the  vanishing  point  of  rays ; 
the  point  m in  which  it  intersects  n n is  the  perspective 
of  a point  of  shadow  on  the  interior  of  the  cylinder. 

To  find  the  shadow  cast  on  any  particular  element,  as 
//",  draw  from  the  vanishing  point  of  horizontal  pro- 
jections a line,  as  H /',  through  its  foot,  and  through  the 
upper  extremity  of  the  element  passing  through  the  other 
point  in  which  H/'  intersects  the  perspective  of  the  base, 
let  a line  be  drawn  to  the  vanishing  of  rays ; the  point 
p,  in  which  it  intersects  the  element  ff  \ is  the  point  of 
shadow  required. 

To  find  the  perspective  of  the  shadow  cast  on  the 
horizontal  plane : 


122 


TREATISE  ON 


The  traces  of  the  tangent  planes  of  rays  are  tangent 
to  the  shadow  cast  by  the  upper  circle  of  the  cylinder 
on  the  horizontal  plane  (27) ; therefore,  their  perspec- 
tives are  tangent  to  the  perspective  of  that  shadow  (88). 
But  the  rays  of  light  passing  through  the  upper  extremi- 
ties of  the  elements  of  shade,  intersect  the  traces  of 
the  tangent  planes  at  the  points  of  tangency  (27); 
therefore,  ¥ and  where  the  perspectives  of  the  rays 
intersect  the  perspectives  of  the  traces,  are  the  perspec- 
tives of  the  points  of  tangency,  and  M",  kk"  are  the  per- 
spectives of  the  shadows  cast  by  the  elements  of  shade. 

That  part  of  the  upper  base  whose  perspective  is 
k'anf'K^  casts  the  curve  of  shadow  on  the  horizontal 
plane ; the  remaining  part  casts  a shadow  on  the  interior 
of  the  cylinder. 

Any  line,  as  Hw^,  drawn  through  H,  may  be  considered 
as  the  perspective  of  the  horizontal  trace  of  a plane 
of  rays ; the  point  n \ in  which  the  perspective  of  the  ray 
through  n intersects  Hw,  is  the  perspective  of  a point 
of  shadow  on  the  horizontal  plane. 

A line  drawn  through  R,  tangent  to  the  perspective 
of  the  upper  base,  will  also  be  tangent  to  the  curve 
kVK\ 

The  part  of  the  surface  of  the  cylinder  which  is  in 
the  shade,  and  seen,  is  shaded  in  the  drawing.  The 
perspective  of  the  shadow  on  the  horizontal  plane  is 
also  shaded. 


LINEAR  PERSPECTIVE. 


123 


PROBLEM  VII. 

It  is  required  to  find  the  perspective  of  the  frustum  of  an 
inverted  cone  ; also  the  perspective  of  the  shadow  on  the  inte- 
rior of  the  frustum^  and  the  perspective  of  the  shadow  on  the 
horizontal  plane » 

93.  Let  the  circle  described  with  the  centre  A and 
radius  AB  (PI.  16),  be  the  horizontal  projection  of  the 
upper  base  of  the  frustum,  and  B'CG'  the  intersection 
of  its  plane  with  the  perspective  plane.  Let  the  circle 
described  with  the  centre  A and  radius  AH,  be  the  lower 
base  of  the  frustum,  and  HI'  its  vertical  projection. 

The  horizontal  projection  of  the  vertex  of  the  cone 
is  at  A,  and  its  vertical  projection,  which  is  at  L,  is  found 
by  joining  B'  and  H'  and  producing  the  line  until  it  in- 
tersects CAE  drawn  perpendicular  to  the  ground  line. 

Let  S be  the  centre  of  the  picture  and  D the  vanish- 
ing-point of  diagonals. 

Through  (E,  C),  a point  of  the  upper  base,  draw  a 
diagonal  and  perpendicular.  The  diagonal  pierces  the 
perspective  plane  at  G',  and  the  perpendicular  pierces  it 
at  C.  The  diagonal  has  its  vanishing  point  at  D,  and 
the  perpendicular  its  vanishing  point  at  S.  Therefore, 
e is  the  perspective  of  the  point  (E,C).  By  similar  con- 
structions we  find  c',  the  perspective  of  (E',C),  c 
the  perspective  of  the  point  whose  horizontal  projec- 
tion is  c,  and  h the  perspective  of  (B,B').  The  perspec- 
tive of  any  point  may  be  found  by  determining  the  per- 
spectives of  the  diagonal  and  perpendicular  passing 
through  it. 

Before  describing  the  ellipse  bece^  it  will  be  well  to 
remember  that  c'S  and  B'S,  being  the  perspectives  of 


124 


TREATISE  ON 


tangent  lines  to  the  upper  base  of  the  frustum,  are  tan- 
gent to  the  ellipse  bece  : and  also,  that  the  tangent  lines 
in  space  drawn  through  the  points  (E,C)  and  (E',C)  are 
parallel  to  the  perspective  plane ; hence,  their  perspec- 
tives are  the  horizontal  lines  drawn  through  e and  e, 
tangent  to  the  ellipse  h e ce. 

The  perspective  of  the  lower  base  of  the  frustum  is 
determined  by  constructions  entirely  similar. 

Let  the  perspective  of  the  vertex  of  the  cone  be  next 
found. 

The  perpendiculai"  through  the  vertex  of  the  cone 
pierces  the  perspective  plane  at  L,  the  vertical  projec- 
tion of  the  vertex ; and  the  diagonal  through  the  vertex 
pierces  the  perspective  plane  atN';  therefore  the  point 
L',  where  LS  intersects  N'D,  is  the  perspective  of  the 
vertex. 

If  through  the  point  of  sight  we  suppose  two  planes 
to  be  drawn  tangent  to  the  frustum  of  the  cone,  the 
traces  of  these  planes,  on  the  perspective  plane,  will 
pass  through  the  perspective  of  the  vertex,  and  will 
limit  the  perspective  of  the  cone;  hence,  the  perspec- 
tives of  the  upper  and  lower  circle  will  be  tangent  to 
these  traces.  Let  these  tangents  be  then  drawn  through 
the  point  L'. 

The  point  of  sight  being  above  the  upper  base  of  the 
frustum,  the  whole  of  that  circle  will  be  seen,  and  there- 
fore its  perspective  is  made  full.  A part  only  of  the 
lower  circle  is  seen,  the  perspective  of  this  part  is 
made  full,  and  is  limited  by  the  tangent  lines  drawn 
through  U. 

To  find  the  shadow  which  the  upper  circle  casts  on 
the  interior  of  the  frustum : 

Let  R be  the  vanishing  point  of  rays,  and  P the 
vanishing  point  of  horizontal  projections. 


LINEAR  PERSPECTIVE 


125 


Since  the  ray  of  light  through  the  vertex  of  the  cone 
is  a line  of  every  plane  of  rays  which  intersects  the  cone 
m right-hned  elements,  the  point  in  which  this  ray 
pierces  the  horizontal  plane  is  common  to  the  horizon- 
tal traces  of  all  such  secant  planes  ; hence,  the  perspec- 
tive of  this  point  is  common  to  the  perspectives  of  all 
the  traces. 

But  the  perspective  of  this  point  is  found  in  the  per- 
spective of  the  ray  through  the  vertex  of  the  cone,  and 
in  the  perspective  of  the  horizontal  projection  of  this 
ray  (82).  Through  R draw  RL' ; this  line  is  the  indefi- 
nite perspective  of  the  ray.  Through-  a',  the  perspec- 
tive of  A,  draw  Pa  ; this  is  the  indefinite  perspective  of 
the  horizontal  projection  of  the  ray;  the  point  K,  in 
which  they  intersect,  is  the  perspective  of  the  point  in 
which  the  ray  through  the  vertex  of  the  cone  pierces 
the  horizontal  plane. 

Through  K,  draw  K/  and  KJ  tangent  to  the  perspec- 
tive of  the  lower  base  of  the  frustum.  These  tangents 
are  the  perspectives  of  the  horizontal  traces  of  the  two 
planes  of  rays  which  are  tangent  to  the  frustum  in  space. 
Hence, L/A  and  ^dg  are  the  perspectives  of  the  elements 
of  shade,  aiid  g and  h the  perspectives  of  the  points  at 
which  the  shadow  on  the  interior  of  the  frustum  begins. 

To  find  points  of  this  shadow,  draw  any  line  through 
K,  as  Kpqs\  which  will  be  the  perspective  of  the  hori- 
zontal trace  of  a secant  plane  of  rays.  Through  the 
points  p and  q draw  the  elements  JJpJc  and  ^qs»  F rom 
k draw  ^R  to  the  vanishing  point  of  rays ; the  point  k' 
in  which  it  intersects  L'^,  is  the  perspective  of  a point 
of  shadow.  The  shadow  on  any  particular  element  is 
found  by  drawing  a line  from  K through  its  foot,  and 
then  drawing  the  perspective  of  a ray  through  the  upper 


126 


TREATISE  ON 


extremity  of  the  element  towards  the  source  of  light,  as 
before. 

The  part  of  the  curve  whose  perspective  ghheh 
casts  a shadow  on  the  interior  of  the  frustum ; and  the 
part  whose  perspective  is  hcseg  casts  a shadow  on  the 
horizontal  plane. 

It  is  now  required  to  find  the  shadow  on  the  horizontal 
plane. 

The  point  whose  perspective  is  casts  a shadow  in 
the  trace  of  the  tangent  plane  of  rays ; but  the  per- 
spective of  the  shadow  is  also  in  the  perspective  of  the 
ray  through  h ; hence  it  is  at  fi.  For  a similar  reason 
g is  the  perspective  of  a point  of  the  shadow  on  the 
horizontal  plane. 

The  line  lS.pqs\  as  has  already  been  remarked,  is  the 
perspective  of  the  trace  of  a secant  plane  of  rays.  If 
through  5,  the  upper  extremity  of  the  element  opposite 
the  source  of  light,  ^ R be  drawn  to  the  vanishing  point 
of  rays,  the  point  s\  in  which  it  meets  Kpy/,  is  another 
point  of  the  perspective  of  the  shadow.  . By  similar  con- 
structions any  number  of  points  may  be  found. 

The  perspective  of  the  shadow  is  tangent  to  the  lines 
KA'  and  K^',  at  the  points  fi  and^'.  A.  line  drawn  through 
R,  tangent  to  perspective  of  the  upper  base  of  the 
frustum  will  also  be  tangent  to  the  curve  gsk'. 

The  part  of  the  exterior  surface  of  the  frustum  which 
is  in  the  shade,  and  seen,  is  shaded  in  the  drawing.  We 
also  shade  that  part  of  the  interior  of  the  frustum  on 
which  the  shadow  falls,  and  which  is  seen.  Of  the 
shadow  which  falls  on  the  horizontal  plane,  all  that  is 
seen  is  shaded. 


LINEAR  PERSPECTIVE. 


127 


PROBLEM  VIII. 

To  find  the  perspective  of  a niche^  and  the  perspective  of  the 
shadow  cast  on  its  interior  surface,  • 

94.  I^et  the  perspective  plane  be  taken  through  the 
front  face  of  the  niche.  Let  AB  (PL  17)  be  the  ground 
line,  and  the  semicircle  bar  the  horizontal  projection  of 
the  niche. 

Draw  b c and  re,  in  the  perspective  plane,  perpen 
dicular  to  the  ground  line,  and  make  them  equal  to  the 
height  of  the  cylindrical  part  of  the  niche;  and  on  ce 
describe  the  semicircle  cle.  These  are  the  lines  of  the 
niche  which  are  in  the  perspective  plane. 

The  perspective  of  the  lower  base  of  the  niche  is 
r a hb,,  the  segment  of  an  ellipse  : the  perspective  of  the 
semicircle  of  contact  of  the  cylindrical  and  spherical 
parts,  is  the  elliptical  segment  cntpe ; both  of  these  curves 
are  found  by  methods  already  explained.  The  lines  bS 
and  rS,  are  the  perspectives  of  the  two  lines  drawn  tan- 
gent to  the  base  of  the  niche  at  the  points  b and  r ; hence 
they  are  tangent  to  the  elliptical  segment  ra'hb.  (88.) 

To  find  the  shadow  on  the  interior  of  the  niche.  The 
first  line  which  casts  a shadow  on  the  interior  of  the 
niche,  is  the  element  c b.  If  through  this  element  a plane 
of  rays  be  passed,  it  will  intersect  the  surface  of  the 
niche,  in  space,  in  a second  element  on  which  the  shadow 
will  fall.  But,  since  the  plane  of  rays  is  vertical,  its 
horizontal  trace  is  parallel  to  the  horizontal  projection 
of  the  rays  of  light ; hence  its  vanishing  point  is  at  H, 
the  vanishing  point  of  horizontal  projections  of  rays, 
and‘6H  is  its  perspective. 


128 


TREATISE  ON 


Through  the  point  in  which  6H  intersects  rdb^  draw 
d (/'parallel  to  6c;  (/(/'  is  the  indefinite  perspective  of  the 
element  that  receives  the  shadow.  Through  c,  the  upper 
extremity  of  the  element  casting  the  shadow,  draw  cR 
to  the  vanishing  point  of  rays ; the  point  d\  in  which  it 
intersects  d d\  is  the  perspective  of  the  shadow  cast  by 
the  point  c,  and  d d'  is  the  shadow  cast  on  the  cylindrical 
part  of  the  niche,  by  the  element  h c.  The  line  bd\%  the 
perspective  of  the  shadow  which  a part  of  the  same 
element  casts  on  the  horizontal  plane. 

Through  H draw  any  line,  as  H/,  near  to  H6.  This 
line  may  be  regarded  as  the  perspective  of  the  horizontal 
trace  of  a vertical  plane  of  rays ; and  h H is  the  per- 
spective of  the  element  in  which  it  intersects  the  cylin- 
drical part  of  the  niche.  The  plane  intersects  the  per- 
spective plane  in  the  line  //' ; therefore  the  point  6',  in 
which/' R intersects  M',  is  the  perspective  of  the  shadow 
cast  by  the  point  (/, /')  on  the  cylindrical  part  of  the  niche 

To  find  the  perspective  of  the  shadow  which  falls  on 
the  spherical  part  of  the  niche : 

If  we  suppose  the  quadrant  of  the  sphere,  which 
forms  the  spherical  part  of  the  niche,  to  be  intersected 
by  a plane  parallel  to  the  front  face  of  the  niche,  the 
section  will  be  a semicircle  whose  diameter  will  be  a 
chord  of  the  semicircle  of  contact  of  the  cylindrical 
and  spherical  parts  of  the  niche.  The  front  circle  of 
the  niche  will  cast  a shadow  on  this  plane  equal  to  itself 
(28) ; and  the  point  where  this  shadow  intersects  the 
circle  cut  from  the  sphere  is  a point  of  the  required 
shadow  in  space.  If  then,  we  find  the  perspectives  of 
these  two  circles,  the  point  in  which  they  intersect  will 
be  the  perspective  of  a point  of  the  curve  of  shadow. 

Since  both  the  circles  are  parallel  to  the  perspective 
plane,  their  perspectives  will  be  circles  (91).  Draw  any 


LINEAR  PERSPECTIVE. 


129 


line,  as  np^  to  represent  the  perspective  of  the  diameter 
of  the  semicircle  cut  from  the  spherical  part  of  the  niche 
by  the  parallel  plane,  and  on  it  describe  the  semicircle 
nkp.  The  perspective  of  the  shadow  cast  on  the  plane 
of  this  circle  by  the  centre  is  found  in  and  also 
in  the  perspective  of  the  projection  of  the  ray  through 
g on  that  plane.  But,  since  the  plane  is  parallel  to  the 
perspective  plane,  the  projection  of  the  ray  upon  it  is 
parallel  to  SR.  Butg*'  is  the  perspective  of  one  point 
of  its  projection ; hence  gg\  drawn  parallel  to  SR,  is  the 
perspective  of  the  projection  of  the  ray  on  the  parallel 
plane,  and  g'  is  the  perspective  of  the  shadow  cast  by 
the  centre  g.  The  shadow  cast  by  eg  is  parallel  to  itself 
and  to  the  perspective  plane ; hence  g'q^  drawn  parallel 
to  eg,  is  the  perspective  of  the  indefinite  shadow  cast  by 
eg  on  the  parallel  plane.  But  this  shadow  is  limited  by 
the  ray  eR ; hence is  the  perspective  of  the  radius  of 
the  circle  of  shadow.  With  g"  as  a centre,  and  the 
radius  gq^  describe  the  arc  qh  ; the  point  in  which  it 
intersects  nkp^  is  the  perspective  of  a point  of  the  curve 
of  shadow. 

If  through  the  centre  g,  gi  be  drawn  in  the  perspec- 
tive plane,  and  perpendicular  to  SR,  it  will  be  a line  of 
the  plane  of  the  circle  of  shadow ; therefore  the  point  e, 
in  which  it  meets  the  circumference  c / c,  is  a point  of 
the  perspective  of  the  curve  of  shadow. 

We  can  easily  find  the  perspective  of  the  point  at 
which  the  shadow  passes  from  the  spherical  to  the 
cylindrical  part  of  the  niche.  For,  if  through  the  point 
in  space  of  which  k is  the  perspective,  a line  be  sup- 
posed drawn  parallel  to  g2,  it  will  be  contained  in  the 
plane  of  shadow,  and  will  pierce  the  upper  base  of  the 
cylinder  in  the  trace  of  the  plane  of  shadow.  But  ks^ 
drawn  parallel  to  gi,  is  the  perspective  of  this  parallel, 


130 


TREATISE  ON 


and  s is  the  perspective  of  the  point  in  Avhich  it  pierces 
the  upper  base  of  the  cylinder.  Hence,  gst  is  the  per- 
spective of  the  trace  of  the  plane  of  shadow,  and  t the 
perspective  of  the  point  at  which  the  shadow  passes 
from  the  spherical  to  the  cylindrical  surface. 

There  are  other  constructions  for  finding  the  shadow 
on  the  spherical  part  of  the  niche. 

Revolve  the  plane  of  rays  passing  through  the 
point  of  sight  and  perpendicular  to  the  perspective 
plane,  about  SR,  until  it  coincides  with  the  perspective 
plane.  The  point  of  sight  falls  in  SD",  drawn  perpen- 
dicular to  SR,  and  at  a distance  from  S equal  to  SD 
(76).  The  ray  of  light  through  the  point  of  sight  takes 
the  position  RD". 

Intersect  the  spherical  part  of  the  niche  by  a plane  of 
rays  perpendicular  to  the  vertical  plane.  Let  Imz, 
drawn  perpendicular  to  gi,  or  parallel  to  SR,  be  the 
trace  of  such  a plane.  Revolve  this  plane  until  it  coin- 
cides with  the  perspective  plane.  The  semicircle  cut 
out  of  the  sphere,  when  revolved,  is  the  semicircle  /mr, 
and  the  ray  through  I takes  the  position  Im^  parallel  to 
RD".  Let  the  counter  revolution  be  now  made,  and 
draw  m'S,  which  is  the  perspective  of  mm',  and  ZR,  which 
is  the  perspective  of  the  ray ; the  point  m",  in  which  they 
intersect,  is  the  perspective  of  a point  of  the  curve  of 
shadow.  Other  points  may  be  found  by  similar  con- 
structions. 


LINEAR  PERSPECTIVE. 


VM 


PROBLEM  IX. 

To  find  the  perspective  of  a sphere^  the  perspective  of  its 
shade^  and  the  perspective  of  its  shadow  on  a horizontal 
plane, 

95.  If  we  suppose  the  sphere  to  be  circumscribed  by 
a visual  cone,  the  curve  in  which  the  perspective  plane 
intersects  the  cone  will  be  the  perspective  of  the  sphere. 
The  point  of  sight  and  the  place  of  the  sphere  being 
given,  the  perspective  plane  may  be  taken  in  such  a 
position  as  to  intersect  the  visual  cone  in  any  of  the 
conic  sections. 

Although  it  is  easy  to  understand  why  the  perspective 
of  a sphere  may  be  an  ellipse,  a parabola,  or  an  hyper- 
bola yet  neither  of  these  curves  seems  to  be  a proper 
representation  of  a body  perfectly  round.  It  is  for  this 
reason  that  the  perspective  of  a sphere  is  generally 
drawn  on  a plane  perpendicular  to  the  line  joining  the 
point  of  sight  and  its  centre ; when  the  perspective  plane 
has  this  position,  it  intersects  the  visual  cone  in  a circle* 

Let  AB  (PI.  18),  be  the  ground  line,  S the  centre  of 
the  picture,  D the  vanishing  point  of  diagonals,  R the 
vanishing  point  of  rays^  and  H the  vanishing  point  of 
horizontal  projections  of  rays. 

Suppose  the  centre  of  the  sphere  to  be  at  a distance 
behind  the  perspective  plane,  equal  to  its  radius.  ^ 

Let  S be  the  projection  of  its  centre,  and  the  circle 
described  with  the  radius  SC  the  projection  of  the 
sphere  on  the  perspective  plane. 

If  through  the  point  of  sight  we  suppose  a plane  to 
be  passed  perpendicular  to  the  horizontal  and  perspec- 
tive planes,  LSa  will  be  its  trace  on  the  perspective 

12 


132 


TREATISE  ON 


plane.  This  plane  will  intersect  the  sphere  in  a great 
circle,  and  the  visual  cone  in  two  elements  which  will 
be  tangent  to  it.  Let  this  plane  be  revolved  about  its 
trace  LSa , mitil  it  coincides  with  the  perspective  plane. 
The  point  of  sight  will  fall  at  D (76),  and  the  centre  of 
the  circle  cut  out  of  the  sphere  at  C.  With  C as  a 
centre,  and  radius  CS,  describe  the  semicircle,  as  in  the 
figure ; and  then  draw  from  D the  tangent  line  Da'a, 
which  will  be,  in  its  revolved  position,  an  element  of  the 
visual  cone.  This  element  pierces  the  perspective  plane 
at  d ; hence,  the  circle  described  with  the  radius  Sa , is 
the  perspective  of  the  sphere. 

It  is  now  required  to  find ’the  perspective  of  the  curve 
of  shade. 

It  will  first  be  necessary  to  find  the  vanishing  line  of 
the  plane  of  the  circle  of  shade  (80). 

Since  the  plane  of  shade  is  perpendicular  to  the  direc- 
tion of  the  light  (34),  the  required  line  is  the  trace  of 
a plane  passing  through  the  point  of  sight  and  perpen- 
dicular to  the  rays. 

Through  the  point  of  sight  let  a plane  of  rays  be 
passed  perpendicular  to  the  perspective  plane ; RSE  is 
its  trace.  Let  this  plane  be  revolved  about  RE,  to  coin- 
cide Vv^ith  the  perspective  plane.  The  point  of  sight 
falls  at  F,  in  a perpendicular  to  RS,  and  at  a distance 
from  S equal  to  SD. 

Since  the  ray  through  the  point  of  sight  pierces  the 
perspective  plane  at  R,  it  will,  after  the  revolution,  take 
the  position  RF.  If  then  FE  be  drawn  perpendicular 
to  RF,  it  will  be  a line  of  the  plane  through  the  point  of 
sight,  and  perpendicular  to  the  direction  of  the  ray,  and 
E will  be  a point  of  its  trace.  But  RE  is  the  projection, 
on  the  perspective  plane,  of  the  ray  passing  through  the 
point  of  sight  (85) : and  since  the  plane  through  the 


LINEAR  PERSPECTIVE. 


133 


point  of  sight  is  perpendicular  to  this  ray,  the  trace  will 
be  perpendicular  to  its  projection.  Therefore,  GET, 
drawn  perpendicular  to  RE,  is  the  vanishing  line  of  the 
plane  of  shade. 

If  we  suppose  two  planes  of  rays  to  be  drawn  tangent 
to  the  visual  cone  which  determines  the  perspective  of 
the  sphere,  they  will  also  be  tangent  to  the  sphere  in 
space,  and  the  points  of  contact  will  be  points  of  the 
curve  of  shade.  The  perspective  of  these  points  of 
contact  will  be  found  in  the  perspective  of  the  sphere, 
and  in  the  traces  of  the  tangent  planes.  Since  the 
tangent  planes  are  planes  of  rays,  they  will  contain  the 
ray  through  the  point  of  sight ; hence  their  traces  will 
pass  through  the  point  R.  But  these  traces  must  also 
be  tangent  to  the  perspective  of  the  sphere ; hence  R6 
and  Rc,  drawn  tangent  to  the  perspective  of  the  sphere, 
are  the  traces  of  the  tangent  planes  of  rays,  and  b and  c 
are  two  points  of  the  perspective  of  the  circle  of  shade. 
And  since  S is  the  perspective  of  the  centre  of  the 
sphere,  the  lines  bSd  and  cSe  are  the  indefinite  perspec- 
tives of  two  diameters  of  the  circle  of  shade. 

Let  us  suppose,  for  a moment,  the  circle  of  shade  to 
be  represented  by  the  circle  dcbq^  Fig.  w,  and  let  bd  and 
c e be  the  diameters  already  referred  to.  The  traces  of 
the  tangent  planes  of  rays  are  the  tangents  nb  and  nc; 
and  since  the  tangent  planes  of  rays  are  perpendicular 
to  the  plane  of  shade,  they  will  intersect  in  a ray  of  light 
perpendicular  to  the  plane  of  the  circle  pbqc  at  w.  If 
through  this  ray  and  the  centre  a,  a plane  be  passed,  its 
trace  nq  will  bisect  the  angle  cab  and  be  parallel  to  the 
lines  b e and  c d^  joining  the  corresponding  extremities  of 
the  diameters  bd  and  ce.  Hence,  the  plane  of  rays 
whose  trace  is  RE,  intersects  the  plane  of  the  circle  of 
shade  in  a line  making  equal  angles  with  the  diameters 


134 


TREATISE  ON 


whose  perspectives  are  bd  and  ce.  But  this  line  is 
parallel  to  the  chords  joining  the  corresponding  ex- 
tremities of  these  diameters,  and  also  to  the  line  FE,  in 
which  the  plane  of  rays  RE  intersects  the  parallel  plane 
through  the  point  of  sight.  Therefore,  E is  the  common 
vanishing  point  of  the  trace  on  the  plane  of  shade,  and 
of  the  parallel  chords  joining  the  corre-sponding  ex- 
tremities of  the  diameters.  Draw  ^E  and  cE.  The 
points  e and  d in  which  they  intersect  cSe  and  6S  J,  are  the 
perspectives  of  the  extremities  of  these  diameters,  and 
consequently,  of  two  more  points  of  the  curve  of  shade. 

To  find  other  points  of  the  curve  of  shade : 

Let  the  plane  passing  through  the  point  of  sight  and 
parallel  to  the  plane  of  shade  be  revolved  about  its  trace 
Gl,  to  coincide  with  the  perspective  plane.  The  point 
of  sight  falls  at  F',  a distance  from  E equal  to  EF. 
Through  this  point  let  any  two  lines,  as  F'G  and  FI,  be 
drawn  at  right  angles  to  each  other,  and  note  the  points 
G and  I in  which  they  meet  the  trace  Gl.  Let  us  now 
consider  the  plane  to  be  revolved  back  to  its  position  in 
space. 

If  through  the  two  extremities  of  any  diameter  of 
the  circle  of  shade,  two  lines  be  drawn  parallel  to  F'G 
and  F'l,  they  will  be  contained  in  the  plane  of  shade, 
and  their  point  of  intersection  on  the  surface  of  the 
sphere  will  be  a point  of  the  circle  of  shade.  But  these 
two  lines  will  have  G and  I for  their  vanishing  points  (71). 
If  therefore,  through  the  points  d and  6,  we  draw  the  lines 
c?G  and  61,  they  will  be  the  perspectives  of  two  lines 
drawn  through  the  extremities  of  a diameter  at  right 
angles  to  each  other,  and  the  point  in  which  they 
intersect,  is  the  perspective  of  a point  of  the  circle  of 
shade.  The  lines  through  the  points  c and  e determine 
|he  point  /. 


LINEAR  PERSPECTIVE 


135 


If  through  /,  we  draw  fSh,  it  will  be  the  indefinite  per- 
spective of  a dianieter  of  the  circle  of  shade.  Through 
c draw  cl.  This  line  is  the  perspective  of  the  chord 
parallel  to  the  chord  whose  perspective  is  fe ; therefore, 
h is  the  perspective  of  the  other  extremity  of  the  diam- 
eter, and  consequently,  the  perspective  of  a point  of  the 
circle  of  shade. 

Having  found  a sufficient  number  of  points  of  the 
curve  of  shade,  let  it  be  described.  Only  the  part  cfgb^ 
which  is  in  front  of  the  circle  of  contact  of  the  visual 
cone  and  sphere,  is  seen. 

It  is  now  required  to  find  the  perspective  of  the  shadow 
cast  on  the  horizontal  plane. 

It  will  first  be  necessary  to  find  the  perspective  of  the 
horizontal  trace  of  the  plane  of  shade. 

Since  this  trace  is  a horizontal  line,  its  vanishing  point 
is  in  the  line  TSH  (74),  and  since  it  is  a line  of  the  plane 
of  shade,  its  vanishing  point  is  in  the  line  GI ; hence  it 
is  at  T. 

It  is  necessary  in  the  next  place  to  find  the  perspective 
of  the  horizontal  projection  of  the  centre  of  the  sphere. 
To  do  this,  lay  off  from  L to  P the  radius  SC  of  the 
sphere;  P is  the  point  at  which  the  diagonal  through  the 
horizontal  projection  of  the  centre  of  the  sphere  pierces 
the  perspective  plane,  and  the  perpendicular  through  the 
same  point  pierces  it  at  L ; hence  i is  the  perspective 
of  the  horizontal  projection  of  the  centre  of  the  sphere. 

We  will  here  premise,  in  order  to  illustrate  what  fol- 
lows, that  when  a line  intersects  the  perpendicular  from 
the  point  of  sight  to  the  perspective  plane,  its  perspecv 
tive  and  its  projection  on  the  perspective  plane  are  the 
same  line : for,  the  visual  plane  which  determines  its 
perspective  is  then  perpendicular  to  the  perspective 


136 


TREATISE  ON 


A second  point  in  the  perspective  of  the  horizonta 
trace  of  the  plane  of  shade  is  found,  by  finding  the  per- 
spective of  the  point  in  which  any  diameter  of  the  circle 
of  shade  pierces  the  horizontal  plane.  To  simplify  the 
construction,  we  will  take  that  diameter  which  is  parallel 
to  the  perspective  plane. 

Since  the  diameter  is  perpendicular  to  the  ray  through 
the  centre  of  the  sphere,  and  since  the  perspectives  of 
the  two  lines  are  the  same  as  their  projections,  it  fol- 
lows, that  their  perspectives  will  be  at  right  angles  to 
each  other  (Des.  Geom.  51).  But  SR  is  the  perspective 
of  the  ray ; hence,  SF  drawn  perpendicular  to  SR,  is  the 
indefinite  perspective  of  the  diameter.  The  projection 
of  this  diameter  on  the  horizontal  plane,  passes  through 
the  horizontal  projection  of  the  centre  of  the  sphere, 
and  is  parallel  to  the  ground  line ; hence,  its  perspective 
passes  through  ^ and  is  parallel  to  AB  (72).  Therefore, 
k is  the  perspective  of  the  point  in  which  the  diameter 
pierces  the  horizontal  plane,  and  consequently,  a point 
in  the  perspective  of  the  horizontal  trace  of  the  plane 
of  shade;  and  T^  is  the  perspective  of  that  trace. 

The  line  is  the  perspective  of  the  horizontal  pro- 
jection of  the  ray  through  the  centre  of  the  sphere,  and 
SR  is  the  perspective  of  the  ray ; hence  p,  their  point 
of  intersection,  is  the  perspective  of  the  shadow  cast  on 
the  horizontal  plane  by  the  centre  of  the  sphere. 

The  shadow  cast  on  the  horizontal  plane  by  any  dia- 
meter of  the  circle  of  shade,  will  pass  through  the  point 
in  which  the  diameter  pierces  the  horizontal  plane,  and 
also  through  the  point  of  which  p is  the  perspective. 
Therefore,  produce  the  diameter  bd  till  it  meets  Tq,  the 
perspective  of  the  horizontal  trace  of  the  plane  of  shade, 
and  from  r draw  rd'pU ; this  line  is  the  perspective 
of  the  indefinite  shadow  cast  by  the  diameter  on  the 


LINEAR  PERSPECTIVE. 


137 


horizontal  plane.  Through  the  extremities  b and  d of 
the  diameter,  draw  lines  to  R;  the  points  V and  d'  are 
points  of  the  perspective  of  the  shadow.  The  diameter 
e c being  produced  to  gives  the  points  c and  e . The 
perspective  of  the  shadow  will  be  tangent  to  the  lines 
R6  and  Rc  at  the  points  b'  and  c . 

96.  We  can  find,  by  a direct  construction,  the  axes  ot 
the  ellipse,  which  is  the  perspective  of  the  circle  of 
shade. 

That  the  figure  may  not  become  too  complicated,  we 
will  make  the  construction  in  Fig.  m,  in  which  the  pro- 
jection and  perspective  of  the  sphere  are  both  repre- 
sented, and  in  which  R is  the  vanishing  point  of  rays. 

If  through  the  axis  of  a scalene  cone,  having  a cir- 
cular base,  a plane  be  passed  perpendicular  to  the  base, 
it  will  divide  the  cone  into  two  symmetrical  parts.  If 
then,  a plane  be  passed  perpendicular  to  the  plane 
through  the  axis,  it  will  intersect  the  cone  in  a curve 
whose  axis  is  the  intersection  of  the  two  planes.  The 
second  axis  of  the  curve  is  the  line  drawn  through  the 
middle  point  of  the  first,  and  perpendicular  to  it. 

The  visual  cone,  which  is  formed  by  drawing  visual 
rays  to  all  the  points  of  the  circle  of  shade,  is  a scalene 
cone  with  a circular  base.  The  plane  of  rays,  whose 
trace  is  SR,  passes  through  the  axis,  is  perpendicu- 
lar to  the  plane  of  shade,  and  also  to  the  perspective 
plane.  Hence,  the  line  RS,  in  which  it  intersects  the 
perspective  plane,  contains  an  axis  of  the  ellipse  in 
which  the  perspective  plane  intersects  the  visual  cone, 
or  an  axis  of  the  ellipse  which  is  the  perspective  of  the 
circle  of  shade.  The  plane,  whose  trace  is  RS,  also 
intersects  the  plane  of  the  circle  of  shade,  in  a diameter 
perpendicular  to  the  direction  of  the  light.  Let  this  plane 
be  revolved  about  RS  to  coincide  with  the  perspective 


138 


TREATISE  ON 


plane.  The  point  of  sight  falls  at  F ; RF  is  the  revolved 
position  of  the  ray,  the  centre  of  the  sphere  falls  at  and 
psq^  drawn  perpendicular  to  RF,  is  the  diameter  of  the 
circle  of  shade.  Through  p and  q draw  Fp  and  F^' ; 
pq  is  evidently  the  perspective  of  pq^  and  is  an  axis  of 
the  ellipse.  Through  a,  the  middle  point  of  p q^  draw 
Yad,  The  chord  of  the  circle  of  shade,  whose  perspec- 
tive is  the  other  axis  of  the  ellipse,  passes  through  the 
point  a,  in  its  true  position  in  space,  and  is  perpendicular 
io  pdq.  But  since  the  circle  of  shade  is  a great  circle 
of  the  sphere,  the  length  of  the ,chord  is  equal  to  cab. 
Let  cab  be  revolved  about  the  middle  point  a,  till  it  has 
the  position  cdb\  parallel  to  /R.  Through  F draw  F c 
and  F6';  the  points  /andc/,  in  which  they  intersect /R, 
limit  the  other  axis  of  the  ellipse.  But  this  axis  must 
pass  through  a,  and  be  perpendicular  to  pq.  There- 
fore, drawing  the  perpendicular  fd\  and  making  ad'  and 
af  equal  to  a/,  or  a of,  we  have  the  second  axis  of  the 
curve. 

PROBLEM  X. 

To  find  the  perspective  of  the  groined  arch  and  the  per  spec-' 
five  of  its  shadows, 

97.  The  arch  is  supported  by  four  pillars,  capped  by 
cornices,  and  standing  on  pedestals ; the  pedestals  are 
placed  in  the  four  angles  of  a square.  The  arch  itself 
is  formed  by  portions  of  two  equal  cylinders. 

Fig.  n (PI.  19)  represents  the  two  projections  of  the 
pedestals,  the  pillars,  the  cornices,  and  the  arch.  On  the 
line  (a5,  a'6'),  joining  the  points  in  which  the  inner  front 
edges  of  the  pillars,  produced,  pierce  the  upper  plane  of 
the  cornices,  let  the  semicircle  dsb'  be  described,  its  plane 


LINEAR  PERSPECTIVE. 


139 


coinciding  with  that  of  the  front  faces  of  the  pillars ; 
and  on  de  let  there  be  described  an  equal  and  parallel 
semicircle. 

If  now  a right  line  be  moved  from  the  position  (ac/,  a ), 
parallel  to  itself,  and  touching  the  two  semicircles,  it  will 
generate  a cylinder  whose  axis  is  (cm,  c),  and  whose 
elements  are  perpendicular  to  the  front  face  of  the 
arch.  On  the  two  lines  gf  and  rc,  joining  the  points  in 
which  the  edges  of  the  pillars  pierce  the  upper  plane  of 
the  cornices,  let  two  equal  and  parallel  semicircles  be 
described,  and  let  a right  line  be  moved  along  them, 
parallel  to  itself,  generating  a semi-cylinder  whose  axis 
zx  is  parallel  to  the  front  face  of  the  arch,  or  perpendi- 
cular to  its  side  faces. 

The  two  cylinders  which  have  been  generated  are 
equal ; their  axes  are  at  right  angles  to  each  other,  and 
the  surfaces  of  the  cylinders  intersect  in  two  equal  semi- 
ellipses, called  the  groins  of  the  arch. 

These  groins  spring  from  the  upper  plane  of  the  cor- 
nices, the  one  from  n to  the  other  from  I to  t.  They 
intersect  each  other  in  a point  of  which  o is  the  hori- 
zontal projection,  and  which  is  at  a distance  above  the 
plane  of  the  cornices  equal  to  the  radius  cs. 

In  the  construction  of  the  arch,  only  a part  of  each 
cylinder  is  used.  The  parts  alonb  and  dqote,  are  formed 
by  the  cylinder  whose  axis  is  cm;  and  the  parts  rloqv  and 
gnotf,  by  the  cylinder  whose  axis  is  zx ; the  other  parts  of 
both  cylinders  are  supposed  to  be  removed. 

If  we  suppose  the  outer  planes  of  the  pillars  to  be 
produced  above  the  upper  plane  of  the  cornices,  they 
will  form  vertical  faces  of  a prism,  whose  horizontal 
sections  are  squares.  We  shall  call  this  prism  the  solid 
part  of  the  arch. 

Let  the  perspective  plane  be  taken  to  coincide  with 


140 


TREATISE  ON 


the  front  faces  of  the  pedestals  and  cornices,  and  let 
AB  be  the  ground  line.  Let  S be  the  centre  of  the 
picture,  D'  and  D the  vanishing  points  of  diagonals,  R 
the  vanishing  point  of  rays,  and  H the  vanishing  point 
of  horizontal  projections.  From  A and  B lay  off  dis- 
tances equal  to  a side  of  the  squares  which  form  the 
bases  of  the  pedestals,  and  on  them  describe  rectangles 
equal  to  the  front  faces  of  the  pedestals.  Then,  let  the 
perspectives  of  the  pedestals  be  found  as  in  Prob.  2. 

From  a lay  off  ab  equal  to  the  distance  which  the 
pedestal  projects  beyond  the  pillar.  From  a draw  dD 
to  the  vanishing  point  of  diagonals,  and  from  b draw 
to  the  vanishing  point  of  perpendiculars ; d is  the  per- 
spective of  the  point  in  which  an  edge  of  the  pillar 
pierces  the  pedestal : and  the  horizontal  line  through  d is 
the  indefinite  perspective  of  the  line  in  which  the  front 
face  of  the  pillar  intersects  the  pedestal. 

From  the  other  extremity  of  the  line  ab  lay  off  a dis- 
tance equal  to  ab^  and  draw  a line  to  S ; the  point  e,  in 
which  it  intersects  de^  is  tne  perspective  of  the  point  in 
which  a second  edge  of  the  pillar  pierces  the  pedestal 
and  the  point  in  which  it  intersects  the  diagonal  aD,  is 
the  perspective  of  the  point  in  which  a third  edge  pierces 
the  pedestal.  Through  this  latter  point  draw  a parallel 
to  de ; the  point  in  which  it  intersects  bS  is  the  perspec- 
tive of  the  point  in  which  the  fourth  edge  of  the  pillar 
pierces  the  pedestal  The  four  vertical  lines  drawn 
through  these  points  are  the  indefinite  perspectives  of  the 
vertical  edges  of  the  pillar.  The  front  and  inner  side 
faces  only  are  seen. 

From  a draw  a vertical  line  m the  perspective  plane, 
and  on  it  lay  off  ac  equal  to  the  distance  between  the 
upper  plane  of  the  pedestals  and  the  lower  plane  of  the 
cornices,  and  through  c draw  an  indefinite  horizontal 


LINEAR  PERSPECTIVE. 


141 


line  cc\  Since  the  front  faces  of  the  cornices  are  in  the 
perspective  plane,  we  can  lay  off  the  thickness  of  the  cor- 
nices c c\  and  their  width  cf^  equal  to  the  width  of  the 
pedestals.  Project  the  point  b into  the  lower  plane  of 
the  cornices  at  b\  and  through  b'  draw  b'S.  The  point  in 
which  this  line  intersects  the  edge  of  the  column  through 
i,  is  the  perspective  of  the  point  in  which  that  edge 
pierces  the  lower  plane  of  the  cornice ; and  the  hori- 
zontal line  through  this  point  is  the  perspective  of  the 
line  in  which  the  front  face  of  the  column  intersects  the 
lower  plane  of  the  cornice.  From  the  point  in  which 
this  horizontal  line  intersects  the  edge  of  the  column 
through  e,  draw  a line  to  S ; this  line  is  the  perspective 
of  the  intersection  of  the  inner  side  face  of  the  column 
with  the  cornice.  The  line  drawn  from  c to  S is  the 
indefinite  perspective  of  the  edge  of  the  cornice  which 
is  perpendicular  to  the  perspective  plane  at  c ; the  part 
which  is  seen  is  limited  by  the  edge  of  the  pillar. 

F rom  /,  draw  /S,  and  from  c,  a diagonal  cD ; their 
point  of  intersection  is  the  perspective  of  the  angle 
of  the  cornice,  diagonally  opposite  to  c.  Through  f‘ 
draw  a vertical  line,  and  from  the  upper  extremity  of  the 
perpendicular  through  draw  a line  to  S ; their  point  of 
intersection  is  the  perspective  of  an  angular  point  of  the 
cornice.  The  horizontal  line  through  f is  the  perspec- 
tive of  the  edge  parallel  to  cf.  We  have  now  found  the 
perspective  of  one  pillar  and  one  cornice.  The  perspec- 
tives of  the  others  are  found  by  similar  constructions. 

We  will  next  find  the  perspectives  of  the  front  and 
back  circles  of  the  arch. 

Find  e and  e",  the  perspectives  of  the  points  in  which 
the  front  and  inner  edges  of  the  pillars  pierce  the  upper 
planes  of  the  cornices ; draw  ee\  and  on  it  describe  a 
semicircle,  making  that  part  full  which  is  above  the 


142 


TREATISE  ON 


upper  lines  of  the  cornices ; this  circle  is  the  perspective 
of  the  front  circle  of  the  arch. 

Find  the  perspectives  of  the  points  in  which  the  edges 
of  the  pillars  that  are  projected  at  d and  e (Fig.  w),  pierce 
the  upper  plane  of  the  cornices ; the  semicircle  described 
on  the  line  joining  these  points,  is  the  perspective  of 
the  back  circle. 

We  next  find  the  perspectives  of  the  groins,  and  the 
perspectives  of  the  side  circles. 

First,  find  the  perspectives  of  the  points  I and  t 
(Fig.  n),  in  the  upper  plane  of  the  cornices,  from  which 
the  groins  spring ; and  also  the  perspectives  of  the  points 
g*,/,  r and  in  the  same  plane,  from  which  the  side 
circles  spring. 

Let  us  now  suppose  the  solid  part  of  the  arch  to  be 
intersected  by  a 'horizontal  plane,  and  let  hi  (Fig.  n),  be 
its  trace  on  the  front  face  of  the  arch.  This  plane  will 
intersect  the  solid  part  of  the  arch  in  a square.  It  will 
intersect  the  cylinder  whose  axis  is  cm,  in  two  elements 
perpendicular  to  the  front  face  of  the  arch  at  the  points 
1c  and  p.  If  through  the  points  in  which  the  horizontal 
projection  of  either  element  intersects  the  lines  nq  and 

lines  be  drawn  parallel  to  the  ground  line,  they  will 
be  the  horizontal  projections  of  the  elements  in  which 
the  secant  plane  intersects  the  other  cylinder.  The 
points  p\  p\  k'  and  k'\  are  in  the  groins,  and  the  square 
Icppk"  has  the  same  diagonals  with  the  square  in  which 
the  secant  plane  intersects  the  solid  part  of  the  arch. 

Draw  any  horizontal  line,  as  hi^  for  the  perspective  of 
the  trace  of  such  a secant  plane.  From  h draw  hS  and 
ht> ; from  i draw  ^S  and  ^D' ; and  from  the  points  k and 
j»,  draw  k^  andpS.  The  points  k'  andp"  are  the  per- 
spectives of  the  points  1c  andj?"  (Fig.  w),  of  the  groin 


LINEAR  PERSPECTIVE. 


143 


It',  and  the  points  p and  k”,  the  perspectives  of  the 
points  p and  ¥ of  the  groin  nq. 

The  Hne  k'p  is  the  perspective  of  the  element  k'p 
(Fig.  n),  and  therefore  the  points  K and  in  which  it  in- 
tersects AS  and  ^S,  are  the  perspectives  of  points  K and 
t (Fig.  w),  of  the  side  circles.  If  we  draw  a line  through 
the  points  ¥ andp",  the  points  A"  and  in  which  it  in- 
tersects AS  and  eS,  are  the  perspectives  of  the  two  points 
A"  and  i"  (Fig.  n),  of  the  side  circles. 

If  we  use  a second  secant  plane,  we  determine  two 
other  points  in  the  perspective  of  each  groin,  and  two 
points  in  the  perspective  of  each  of  the  side  circles. 

If  through  the  point  of  sight  a plane  be  passed  tan 
gent  to  the  cylinder  whose  elements  are  parallel  to  the 
perspective  plane,  it  is  evident  that  the  perspective  of 
the  element  of  contact  will  pass  through  the  highest 
point  of  each  of  the  curves.  But  the  perspective  of 
this  element  is  a horizontal  line  (72) ; hence  it  will  be 
tangent  to  each  of  the  curves. 

To  find  this  element,  and  its  perspective,  suppose  a 
vertical  plane  to  be  passed  through  the  point  of  sight, 
and  perpendicular  to  the  perspective  plane.  This  plane, 
whose  trace  is  R'^,  will  intersect  the  cylinder  in  a circle 
whose  centre  is  in  the  upper  plane  of  the  cornices,  and 
equally  distant  from  the  front  and  back  faces  of  the 
arch.  Let  this  plane  be  revolved  about  R'^,  to  coincide 
with  the  perspective  plane.  The  point  of  sight  falls  at 
D,  and  the  centre  of  the  circle  at  c.  With  c as  a 
centre,  and  ca,  equal  to  a line  corresponding  to  ca 
(Fig.  n),  describe  the  arc  of  a circle,  and  from  D draw  a 
tangent  to  this  arc.  This  tangent  is  a line  of  the  tan- 
gent plane,  and  the  point  in  which  it  meets  S^,  produced, 
is  the  point  at  which  it  pierces  the  perspective  plane.  T he 
horizontal  line  drawn  through  this  point  is  the  perspec- 


144 


TREATISE  ON 


tive  of  the  element  to  which  the  curves  are  tangent. 
Let  the  curves  be  now  described. 

The  points  at  which  the  perspectives  of  the  groins 
intersect  the  perspectives  of  the  side  circles,  will  limit 
the  parts  of  the  side  circles  which  are  seen. 

It  is  now  required  to  find  the  perspective  of  the 
shadows  cast  by  the  different  parts  of  the  arch. 

The  front  circle  of  the  arch  casts  a shadow  on  that 
part  of  the  arch  formed  by  the  cylinder  whose  elements 
are  perpendicular  to  the  perspective  plane.  In  finding 
the  perspective  pf  this  shadow,  we  have  merely  to  find 
the  perspective  of  the  shadow  cast  by  the  base  of  a 
cylinder  on  the  interior  surface. 

Throuhg  g,  draw  gl  perpendicular  to  SR ; I is  the  per- 
spective of  the  point  at  which  the  shadow  on  the  interior 
of  the  cylinder  begins ; for  the  tangent  line  at  I is  parallel 
to  SR,  and  is  the  perspective  of  the  trace  of  a plane  of 
rays  tangent  to  the  cylinder.  To  find  other  points  of 
the  shadow,  draw  any  line,  as  mn^  parallel  to  the  tangent 
at  L This  line  is  the  perspective  of  the  trace  of  a plane 
of  rays  on  the  front  face  of  the  arch.  This  plane  inter- 
sects the  cylinder  in  two  elements ; one  casts,  and  the 
other  receives  the  shadow.  The  line  wS  is  the  perspec- 
tive of  the  element  which  receives  the  shadow,  and  m\ 
where  it  is  intersected  by  mR,  is  the  point  of  shadow. 
Let  the  curve  of  shadow  from  I be  then  described. 

The  rays  of  light  may  have  such  a direction,  that  the 
shadow  of  the  front  circle  will  intersect  the  groin  k'p" 
When  this  occurs,  one  curve  of  shadow  will  pass  from 
the  point  of  intersection  along  the  cylinder  whose 
elements  are  parallel  to  the  perspective  plane,  and  will 
pass  off  on  the  side  circle  ti\  Another  curve  will  pass 
from  the  point  of  intersection,  along  the  cylinder  whose 


LINEAR  PERSPECTIVE.  145 

elements  are  perpendicular  to  the  perspective  plane,  and 
will  pass  off  on  the  back  circle  of  the  arch. 

The  side  circle  of  the  arch,  on  the  left,  casts  a shadow 
on  the  interior  of  the  arch  which  falls  on  the  cylinder 
whose  elements  are  parallel  to  the  perspective  plane. 
If  w^e  suppose  a tangent  plane  of  rays  to  be  drawn  to 
this  cylinder,  the  point  in  which  the  element  of  contact 
meets  the  end  circle  towards  the  source  of  light,  will  be 
the  point  at  which  the  shadow  on  the  interior  of  the 
cylinder  begins.  But  this  tangent  plane  will  be  perpen- 
dicular to  the  side  face  of  the  arch ; hence,  its  trace 
will  be  parallel  to  the  projections  of  rays  on  the  side 
planes.  If  we  suppose  a plane  to  be  passed  through  R 
parallel  to  the  side  planes  of  the  arch,  and  the  ray 
through  the  point  of  sight  to  be  projected  on  this  plane, 
the  projection  will  be  parallel  to  the  projection  of  rays 
on  side  planes,  and  consequently,  to  the  trace  of  the 
tangent  plane.  If  a line  be  drawn  through  the  point  of 
sight  parallel  to  the  projection  of  the  ray,  it  will  pierce 
the  perspective  plane  at  R,  since  RR'  is  the  trace,  on  the 
perspective  plane,  of  the  plane  which  projects  the  ray  on 
the  plane  through  R ; hence  R'  is  the  vanishing  point  of 
the  projections  of  rays  on  the  side  planes. 

If,  therefore,  through  R we  draw  R^^ , tangent  to  the 
curve  KqU\  it  will  be  the  perspective  of  the  trace  of  the 
tangent  plane  of  rays,  and  is  the  perspective  of  the  point 
at  which  the  shadow  on  the  interior  of  the  arch  begins. 

If  through  R we  draw  a secant  line,  we  may  regard  it 
as  the  perspective  of  the  trace  of  a plane  of  rays,  paral- 
lel to  the  tangent  plane,  and  intersecting  the  cylinder  in 
two  elements. 

If  through  the  lower  point,  in  which  this  line  cuts  the 
curve  h'qh",  we  draw  a horizontal  line,  it  will  be  the  per- 
spective of  the  element  which  receives  the  shadow ; and 

K 


146 


TREATISE  ON 


drawing  through  the  upper  point  a line  to  the  vanishing 
point  of  rays,  determines  o,  the  perspective  of  a point 
of  the  curve  of  shadow.  Let  the  curve  of  shadow  be 
then  drawn. 

We  are  next  to  find  the  shadows  of  the  cornices  on 
the  front  faces  of  the  pillars. 

Through  c draw  a line  to  S,  and  another  to  R.  The 
point  where  the  line  to  S meets  the  line  in  which  the  front 
face  of  the  pillar  intersects  the  lower  plane  of  the  cor- 
nice, is  the  perspective  of  one  point  in  the  projection  of 
■the  ray  through  c,  on  the  front  face  of  the  pillar,  and  the 
line  drawn  through  this  point,  parallel  to  SR,  is  the  per- 
spective of  the  projection : the  point  in  which  this 
perspective  intersects  cR  is  the  perspective  of  one  point 
of  the  line  of  shadow : but  the  line  of  shadow  is  hori- 
zontal both  in  space  and  in  perspective. 

To  find  the  shadow  cast  by  the  column.  The  edge  e r 
casts  a line  of  shadow  on  the  pedestal  and  on  the  hori- 
zontal plane,  both  of  which  are  parallel  to  the  horizontal 
projection  of  rays.  Hence,  their  perspectives  are  easily 
found.  At  s the  shadow  falls  on  the  inner  side  face  of 
the  back  pedestal.  It  passes  up  the  face  in  a vertical 
line ; reaching  the  upper  face  it  becomes  parallel  to  the 
horizontal  projections  of  rays,  and  when  it  reaches  the 
face  of  the  pillar,  it  ascends  in  a vertical  line  along  the 
face.  If  through  r,  the  highest  point  of  the  edge  of  the 
pillar  which  is  in  the  light,  a line  be  drawn  to  R,  it  will 
limit  the  shadow  cast  by  the  edge  of  the  pillar.  F rom 
the  point  in  which  this  line  intersects  the  vertical  line 
before  drawn,  on  the  face  of  the  pillar,  the  shadow  will 
be  cast  by  the  lower  line  of  the  cornice.  This  shadow 
will  be  parallel  to  the  projections  of  rays  on  the  side 
planes,  will  therefore  have  its  vanishing  point  at  R',  and 
will  be  limited  by  the  ray  through  /.  The  vertical  line 


LINEAR  PERSPECTIVE. 


147 


through  f will  then  cast  its  shadow,  which  will  be  a 
vertical  line,  and  will  be  limited  by  the  ray  drawn 
through  the  upper  extremity  of  the  vertical  edge  of  the 
cornice  through  f.  There  is  a small  part  of  the  edge 
of  the  cornice,  parallel  to  //',  which  is  in  the  light,  and 
which  will  cast  a shadow  on  the  face  of  the  pillar.  Its 
shadow  will  be  perpendicular  to  the  perspective  plane, 
and  consequently  have  its  vanishing  point  at  S. 

The  front  circle  of  the  arch  will  next  cast  a shadow 
on  the  side  face. 

To  find  this  shadow  we  will  observe,  that  the  shadow 
of  the  diameter  ee\  on  the  side  face,  is  parallel  to  the 
projections  of  rays  on  the  side  face,  and  since  e is  the 
perspective  of  the  point  in  which  the  diameter  pierces 
the  plane  of  the  side  face,  will  be  the  indefinite  per- 

spective of  its  shadow.  Draw  any  ordinate  of  the  cir- 
cle, as  /M,  which  it  is  supposed  will  cast  a shadow. 
Through  t draw  ; the  point  t is  the  shadow  cast  on 
the  side  plane  by  the  foot  of  the  ordinate.  But  the  ordi- 
nate being  parallel  to  the  side  plane,  its  shadow  will  be 
the  vertical  line  through  t\  The  point  in  which  this 
vertical  line  intersects  the  ray  through  ii  is  a point  of 
shadow  on  the  side  plane.  When  this  shadow  reaches 
the  front  face  of  the  pillar,  it  then  falls  on  that  face,  and 
is  a circle  both  in  space  and  in  perspective. 

Through^  draw  ^R ; and  through  g\  the  perspective 
of  the  point  at  which  the  centre  of  the  front  circle  is 
projected  on  the  plane  of  the  front  faces  of  the  back 
pillars,  draw  gg”  parallel  to  SR ; g is  the  perspective 
of  the  shadow  cast  by  the  centre  of  the  front  circle 
on  the  plane  of  the  front  faces  of  the  back  pillars 
But  the  shadow  cast  by  the  radius  ge  is  parallel  to  itself ; 
therefore,  draw  ge"  parallel  to  the  ground  line,  and  e"R 
to  the  vanishing  point  of  rays ; g"  e"  is  the  perspective 

K2 


148 


TREATISE  ON 


of  the  radius  of  the  circle  of  shadow.  With  g"  as  a 
centre,  and  g''  d"  as  a radius,  describe  the  arc  of  a circle, 
and  we  have  the  shadow  cast  on  the  pillar.  The  parts 
of  the  arch  which  are  in  the  shade,  or  on  which  shadows 
fall,  are  shaded. 

PROBLEM  XL 

It  is  required  to  find  the  perspective  of  a house  and  the  perspec-^ 
live  of  its  shadows. 

98.  Having  measured  ali  the  lines  of  the  house  which 
are  to  be  put  in  perspective,  make  its  horizontal  pro- 
jection to  any  convenient  scale,  as  in  PL  20. 

The  outer  lines  are  the  horizontal  projections  of  the 
outer  lines  of  the  eave-trough,  and  the  adjacent  inner 
ones  are  the  lines  in  which  the  outer  faces  of  the  walls 
intersect  the  horizontal  plane.  The  lines  in  which  the 
roofs  intersect  are  also  made,  as  well  as  the  projections 
of  the  chimneys  and  steps. 

In  selecting  the  position  of  the  perspective  plane, 
reference  should  be  had  to  the  part  of  the  building 
which  is  to  appear  most  prominent  in  the  picture.  Were 
it  only  required  to  represent  the  front  of  the  building, 
we  should  take  the  perspective  plane  parallel  to  it ; but 
when  the  front  and  an  end  are  to  be  represented,  it  is 
most  convenient  to  take  the  perspective  plane  oblique  to 
them  both ; though  it  may  be  taken  parallel  to  the  one 
and  perpendicular  to  the  other.  It  simplifies  the  con- 
struction, without  affecting  the  generality  of  the  method, 
*to  assume  the  perspective  plane  through  one  or  two  of 
the  prominent  vertical  lines.  Such  lines  being  in  the 
perspective  plane  will  be  their  own  perspectives,  and  alt 
the  vertical  distances  may  be  laid  off  on  them. 

In  the  construction  here  given,  the  perspective  plane 
is  passed  through  the  vertical  corners  of  the  house 


LINEAR  PERSPECTIVE. 


149 


which  pierce  the  horizontal  plane  at  B and  A,  and  DE 
is  the  ground  line.  Having  chosen  the  position  of  the 
eye,  lay  off  the  distance  FS  equal  to  its  height  above 
the  ground,  and  draw  KSH,  which  will  be  the  horizon 
of  the  picture  (74).  On  the  line  FG,  perpendicular  to 
DE,  lay  off  the  distance  of  the  point  of  sight  from  the 
perspective  plane.  The  horizontal  projection  of  the 
point  of  sight  does  not  fall  on  the  paper,  the  eye  being 
at  about  one  and  a half  times  the  length  of  the  building 
from  the  perspective  plane.  In  making  the  drawing, 
a paper  must  be  taken  of  such  dimensions  as  will  enable 
us  to  represent  upon  it  the  horizontal  projection  of  the 
point  of  sight,  and  the  vanishing  points  of  the  principal 
lines. 

Although  the  horizontal  projection  of  the  house  is 
made  in  front  of  the  ground  line,  yet  the  building  is  in 
fact  behind  the  perspective  plane,  with  its  front  towards 
the  eye.  To  represent  to  the  mind  the  horizontal  pro- 
jection of  the  house  in  its  true  position,  we  must  con- 
ceive the  horizontal  plane  to  be  revolved  180°  around 
the  ground  line  DE,  which  will  bring  every  point  now 
in  front  of  the  ground  line  at  an  equal  distance  be- 
hind it. 

At  the  points  a and  d draw  aha  and  dcd'  perpendi- 
cular to  the  ground  line  DE,  and  make  b a equal  to  a b^ 
and  c d'  equal  to  c d.  The  lines  Kd  and  Ac/',  are  the  lines 
Aa  and  Ac/,  in  the  positions  from  which  their  perspectives 
are  taken. 

Through  the  point  of  sight  let  a line  be  drawn  paral- 
lel to  Ac/' : its  horizontal  projection  will  pass  through 
the  horizontal  projection  of  the  point  of  sight,  and  the 
line  will  pierce  the  perspective  plane  in  a point  of  the 
horizontal  line  KSH.  This  point  is  the  vanishing  point 
of  all  lines  parallel  to  Ac/.  The  point  is  not  on  the 
paper,  but  as  we  have  frequent  occasion  to  refer  to  it, 


150 


TREATISE  0>* 


we  will  designate  it  by  the  letter  K.  Through  the  point 
of  sight  let  a line  be  drawn  parallel  to  Ad : this  line 
pierces  the  perspective  plane  at  H,  which  is,  therefore, 
the  vanishing  point  of  all  lines  parallel  to  Aa. 

From  A draw  a line  to  the  vanishing  point  K;  it  will 
be  the  indefinite  perspective  of  the  line  Ac?.  From  e 
draw  a perpendicular  to  the  ground  line,  and  from  the 
foot  of  the  perpendicular  draw  a line  to  S ; the  point  c , 
in  which  it  intersects  the  line  AK,  is  one  point  in  the  per- 
spective of  the  corner  of  the  house,  which  pierces  the 
horizontal  plane  at  e;  the  vertical  line  drawn  through 
the  point  e is  the  indefinite  perspective  of  the  corner, 
and  the  line  drawn  to  B is  the  perspective  of  the 
line  Be. 

The  line  drawn  from  B to  the  vanishing  point  K,  is 
the  indefinite  perspective  of  the  line  B/.  From  /,  draw 
a perpendicular  to  the  ground  line,  and  from  the  foot  of 
the  perpendicular  draw  a line  to  S ; the  point  f in  which 
it  meets  the  line  BK,  before  drawn,  is  the  perspective  of 
the  point  /.  Through  the  point  f\  draw  a vertical  line, 
and  it  will  be  the  indefinite  perspective  of  the  corner  of 
the  house  which  pierces  the  horizontal  plane  at  f.  The 
perspective  of  the  point  d is  in  the  line  AK,  and  also  in 
the  line  joining  c and  S ; hence  it  is  at  d'  their  point  of 
intersection.  The  vertical  line  drawn  through  this  point 
is  the  indefinite  perspective  of  the  corner  of  the  build- 
ing which  meets  the  horizontal  plane  at  d. 

On  the  vertical  lines  passing  through  the  points  A and 
B,  lay  off  the  distance  from  the  ground  to  the  eave- 
trough.  From  the  upper  extremity  of  the  vertical  line 
through  A,  draw  a line  to  the  vanishing  point  K,  and 
note  the  point  in  which  it  intersects  the  vertical  line 
through  e : the  part  intercepted  between  the  vertical 
lines  through  A and  c,  is  the  perspective  of  the  lower 


LINEAR  PERSPECTIVE. 


151 


line  of  the  eave-trougli  that  is  horizontally  projected  in 
the  line  Ae.  The  line  joining  the  upper  extremities  of 
the  vertical  lines  through  B and  is  the  perspective  of 
the  lower  line  of  the  eave-trough  of  which  Be  is  the 
projection,  and  the  perspective  would,  if  produced,  pass 
through  H.  From  the  upper  extremity  of  the  vertical 
line  through  B,  draw  a line  to  K.  The  part  cut  off  by 
the  vertical  line  through  f'  is  the  perspective  of  the  lower 
line  of  the  eave-trough,  of  which  B/  is  the  horizontal 
projection.  Through  the  upper  extremity  of  the  vertical 
line  through  A,  a line  has  been  drawn  to  K ; the  part  of 
this  line  intercepted  between  the  vertical  lines  through  f' 
and  d\  is  the  perspective  of  all  that  can  be  seen  of  the 
lower  line  of  the  eave-trough,  whose  horizontal  projec- 
tion isgJ. 

From  A draw  a line  to  H,  and  from  b a line  to  S. 
The  vertical  line  drawn  through  d\  their  point  of  inter- 
section, is  the  indefinite  perspective  of  the  corner  of  the 
house  which  meets  the  horizontal  plane  at  a.  From  the 
upper  extremity  of  the  vertical  line  through  A,  draw  a 
line  to  H.  The  part  cut  off  by  the  vertical  line  through 
a is  the  perspective  of  the  lower  line  of  the  eave-trough 
of  which  Aa  is  the  horizontal  projection. 

On  the  vertical  lines  through  A and  B,  lay  off  the  dis- 
tance to  the  lower  line  of  the  water-table ; then  the 
width  of  the  water-table ; then  find  the  perspectives  of 
its  upper  and  lower  lines,  in  the  same  manner  as  we  have 
already  found  the  perspectives  of  the  lines  of  the  eave- 
trough.  On  the  vertical  line  through  A,]ay  off  from  the  line 
DE,  the  distance  to  the  upper  face  of  the  lower  window- 
sill, and  draw  through  the  point  a horizontal  line,  and 
also  a line  to  K.  The  horizontal  line  is  the  projection 
on  the  perspective  plane  of  the  lower  line  of  the  windows 
of  the  first  story,  and  the  line  to  K is  its  perspective. 


152 


TREATISE  ON 


From  the  point  in  which  the  perspective  meets  the  ver 
tical  line  through  c , draw  a line  to  H,  and  it  will  be  the 
perspective  of  the  lower  line  of  the  window,  in  that  part 
of  the  building  whose  horizontal  projection  is  Be : this  line 
intersects  the  vertical  line  through  B,  at  the  same  point 
in  which -it  is  intersected  by  the  projection  of  the  lower 
line  of  the  windows.  The  line  drawn  from  this  point  to 
K is  the  indefinite  perspective  of  the  lower  line  of  the 
windows,  corresponding  to  the  part  of  the  house  of  which 
Bf  is  the  horizontal  projection.  By  similar  constructions 
we  can  find  the  perspectives  of  the  upper  horizontal 
lines  of  the  windows,  and  also  the  perspectives  of  the  hori- 
zontal lines  which  are  tangent  to  their  semicircular  arches. 

In  finding  the  perspectives  of  the  vertical  lines  of  the 
windows,  it  should  not  be  forgotten  that  they  are  paral- 
lel to  the  perspective  plane,  and  consequently,  their 
perspectives  will  be  parallel  to  the  lines  themselves. 
Through  the  lower  extremity  of  the  vertical  line  whose 
horizontal  projection  is  A,  draw  a line  perpendicular  to 
the  perspective  plane.  It  will  pierce  the  perspective 
plane  in  the  projection  of  the  lower  line  of  the  windows. 
From  the  point  at  which  it  pierces,  draw  a line  to  S : 
its  intersection  with  the  perspective  of  the  horizontal 
line  before  referred  to,  is  the  perspective  of  one  point  of 
the  vertical  bounding  line  of  the  window  ; and  the  inde- 
finite line  can  be  drawn,  since  it  is  vertical.  The  other 
vertical  lines  are  found  by  constructions  entirely  similar, 
and  all  of  them  are  limited  by  the  perspectives,  which 
have  already  been  found,  of  the  horizontal  lines  of  the 
building. 

To  find  the  perspective  of  the  roofs  of  the  house. 
Produce  wm,  the  horizontal  projection  of  the  line  in 
which  the  side  roofs  intersect,  until  it  meets  the  ground 
line  at  E.  Draw  EE'  perpendicular  to  the  ground  line, 


LINEAR  PERSPECTIVE. 


153 


and  make  it  equal  to  the  height  of  the  intersection  of 
the  roofs  above  the  ground.  Then  E'  will  be  the  point 
in  which  the  intersection  of  the  roofs  pierces  the  per- 
spective plane-— the  horizontal  line  through  E'  is  the 
projection,  on  the  perspective  plane,  of  the  intersection 
of  the  roofs,  and  the  line  drawn  from  E'  to  the  vanish- 
ing point  K'  is  its  indefinite  perspective.  The  point  of 
which  k is  the  horizontal  projection  is  vertically  pro- 
jected at  k\  From  k'  draw  a line  to  S : the  point  in 
Avhich  it  intersects  the  line  E'K,  is  the  perspective  of  the 
point  (^,  ^').  The  perspectives  of  the  points  (m,  m'), 
(w, n')^  and  are  found  by  similar  constructions. 

If  through  the  point  in  which  the  vertical  line  through 
A meets  the  horizontal  plane  of  the  eaves  of  the  house, 
a line  be  drawn  to  k\  it  will  be  the  projection,  on  the 
perspective  plane, of  the  line  in  which  the  side  and  front- 
end  roofs,  intersect. 

The  horizontal  lines  of  the  water-table,  of  the  windows, 
and  of  the  eave-trough,  corresponding  to  the  end  of  the 
house,  have  a common  vanishing  point  H ; and  the  per- 
spectives of  the  vertical  lines  of  the  windows  are  found 
in  the  same  manner  as  those  in  the  front  of  the  house. 

*To  find  the  perspectives  of  the  chimneys.  From  the 
horizontal  line  w E'  lay  off  a distance  above  it  equal  to 
the  height  of  the  tops  of  the  chimneys  above  the  inter- 
section of  the  side  roofs,  and  through  the  point  so  de- 
termined, draw  the  horizontal  line  rs.  Produce  the  hori- 
zontal lines  of  the  cornices  which  intersect  at  either  of 
the  angles  that  are  seen,  until  they  meet  the  perspective 
plane,  which  they  will  do  in  the  line  rs.  Since  H and  K 
are  the  vanishing  points  of  these  lines,  their  perspectives 
can  be  drawn,  and  the  point  in  which  any  two  of  them, 
that  pass  through  the  same  angular  point  of  the  cornice, 
intersect,  is  the  perspective  of  an  angular  point  of  the 


154 


TREATISE  ON 


cornice.  The  perspectives  of  the  lines  in  which  the 
faces  of  the  chimneys  intersect  the  roof  of  the  house, and 
the  perspectives  of  their  vertical  edges,  are  easily  found. 

99.  Although  the  perspective  of  every  point  can  be 
found  rigorously,  yet  the  smaller  parts  of  the  building 
can  generally  be  made  with  sufficient  accuracy,  without 
making  a separate  construction  for  each  of  them. 
Thus,  after  we  have  found  the  bounding  lines  of  the 
windows,  the  window-sills,  the  caps  and  casings  may  be 
made  very  accurately  without  the  aid  of  a geometrical 
construction.  Having  found  the  perspectives  of  the 
doors  and  steps,  we  may  make  the  railing  without  a par- 
ticular construction  for  each  vertical  bar.  After  having 
found  the  lower  line  of  the  eave-trough,  we  may  draw 
the  outer  upper  line,  always  observing  the  rules  of  per- 
spective, and  the  general  symmetry  of  the  picture. 

In  looking  obliquely  upon  a building,  the  casings  of 
the  doors  and  windows  on  the  side  nearest  the  eye  are' 
not  seen,  while  those  on  the  other  side  are  very  distinct. 

To  find  the  perspective  of  the  shadows  cast  upon 
the  house.  Let  R be  the  projection  of  a ray  of  light 
on  the  horizontal  plane,  R'  its  projection  on  the  perspec- 
tive plane,  and  R"  the  position  of  its  horizontal  projec- 
tion, after  it  has  been  revolved  to  correspond  with  the 
horizontal  projection  of  the  building.  The  rays  of  light 
being  nearly  parallel  to  the  perspective  plane,  their 
vanishing  point  will  be  so  far  distant  from  the  centre  of 
the  picture  that  we  cannot  use  it  conveniently  in  finding 
the  shadows.  We  therefore  adopt  other  methods. 

In  all  the  constructions,  we  must  bear  in  mind  that 
the  house  is  behind  the  perspective  plane,  and  there- 
fore, when  a ray  of  light  is  drawn  through  any  point, 
its  horizontal  projection  must  not  be  drawn  parallel  to 
R,  but  to  R',  which  makes  the  same  angle  with  the 


LINEAR  PERSPECTIVE. 


15b 


ground  line,  but  is  differently  inclined.  Let  us  first  find 
the  shadow  cast  by  the  upper  line  of  the  end  eave- 
trough,  on  the  end  of  the  house. 

From  t draw  a line  perpendicular  to  the  ground  line, 
and  make  v t'  equal  to  the  distance  from  the  ground  to 
the  upper  line  of  the  eave-trough.  Through  (/,  f)  draw 
a ray  of  light ; it  pierces  the  end  wall  of  the  house  in 
the  point  (w,  w'),  which  is  a point  in  the  line  of  shadow. 
From  the  point  in  which  uu  meets  the  ground  line, 
draw  a line  to  S ; the  point  in  which  it  intersects  AH  is 
the  perspective  of  the  point  u.  Through  this  point  draw 
a vertical  line,  and  it  will  be  the  perspective  of  the  vertical 
line  passing  through  {u^  ;/).  From  ?/,  draw  a line  to 
S;  the  point  in  which  it  intersects  the  vertical  line 
before  found,  is  the  perspective  of  one  point  of  the 
required  line  of  shadow.  But  since  the  shadow  is 
parallel  to  the  line  itself,  its  vanishing  point  is  at  H ; 
therefore,  the  line  drawn  through  H,  and  the  point  found, 
is  the  indefinite  perspective  of  the  line  of  shadow. 

Through  the  highest  point,  which  is  in  the  light,  of 
the  corner  of  the  house  which  meets  the  horizontal 
plane  at  A,  draw  a ray  of  light.  Such  ray  will  pierce 
the  face  Be  of  the  building  in  a point  which  is  horizon- 
tally projected  at  i.  The  vertical  line  drawn  through  i 
is  the  shadow  cast  on  the  wall  by  the  corner  of  the 
house  which  meets  the  horizontal  plane  at  A.  The 
perspective  of  this  vertical  line  of  shadow,  which  is 
easily  found,  is  the  perspective  of  the  shadow  cast  by 
the  corner  of  the  house  towards  the  source  of  light. 
From  the  upper  extremity  of  this  shadow,  the  shadow 
is  cast  for  a short  distance  by  the  upper  line  of  the  end 
eave-trough ; then  by  the  upper  line  of  the  front  eave- 
trough;  and  the  shadow  terminates  in  the  line  of 
shadow  cast  by  the  upper  line  of  the  eave-trough  belong- 


156 


TREATISE  ON 


ing  to  the  part  of  the  house  Be ; the  latter  shadow  being 
parallel  in  space  to  the  shadow  cast  by  the  end  eave- 
trough  on  the  end  of  the  house. 

The  shadow  of  one  of  the  chimneys  only  is  seen. 
This  shadow  is  found  by  drawing  rays  of  light  through 
the  extremities  of  the  lines  casting  it ; finding  the  points 
in  which  they  pierce  the  roof,  and  determining  the  per- 
spectives of  those  points 

GENERAL  REMARKS. 

100.  It  is,  perhaps,  too  obvious  to  require  illustration, 
that  the  projections  in  Descriptive  Geometry,  viz.  the 
Orthographic  and  Stereographic  projections  of  the 
sphere,  are  but  particular  methods  of  perspective. 

If  the  point  of  sight  be  at  an  infinite  distance  from 
the  perspective  plane,  the'  perspectives  of  objects  are 
the  same  as  their  orthographic  projections. 

101.  If  the  plane  of  any  circle  of  the  sphere  be  as- 
sumed for  the  perspective  plane,  the  point  of  sight  being 
at  the  pole,  the  perspective  representation  of  the  sphere 
upon  this  plane  will  in  nowise  differ  from  its  stereographic 
projection. 

102.  We  have  thus  far  represented  objects  on  plane 
surfaces  only.  Their  perspectives  may  however  be 
made  on  other  surfaces,  and  the  principles  which  have 
been  explained  will  require  but  a slight  modification  to 
render  them  applicable  to  the  case  in  which  the  per- 
spective is  made  on  any  surface  whatever. 

Suppose,  for  example,  it  were  required  to  construct 
the  perspectives  of  the  different  circles  of  the  earth  on 
the  surface  of  a tangent  cylinder,  the  point  of  sight 
being  at  the  centre.  If  we  suppose  the  equator  to  be 
the  circle  of  contact,  the  perspectives  of  the  meridians 
will  be  elements  of  the  cylinder,  and  the  perspective 


LINEAR  PERSPECTIVE. 


157 


of  any  other  circle  is  found  by  constructing  the  inter- 
section of  the  cone,  of  which  the  circle  is  the  base,  and 
the  point  of  sight  the  vertex,  with  the  surface  of  the 
tangent  cylinder.  Having  thus  found  the  perspectives 
of  all  the  circles  of  the  sphere,  if  we  develop  the  surface 
of  the  tangent  cylinder  on  a plane,  its  development  will 
be  a map  of  the  earth.  Mercator’s  chart  is  constructed 
on  these  principles. 

103.  Panoramic  views,  which  often  exhibit  entire 
cities,  are  generally  constructed  on  the  surface  of  a 
vertical  cylinder,  the.  eye  of  the  spectator  being  in  the 
axis.  When  the  perspective  is  accurately  made,  and 
viewed  from  the  right  point,  the  deception  is  perfect. 
The  houses  seem  to  stand  out  from  the  canvass  on 
which  they  are  drawn ; the  streets  have  the  aspect  of 
bustle  and  business,  and  one  feels  himself  transported 
into  a populous  city,  and  mingling  in  its  affairs. 

104.  Jt  is  of  great  importance  in  linear  perspective, 
that  the  distance  of  the  eye  from  the  plane  of  the  picture 
should  be  judiciously  chosen.  If  that  distance  is  not 
sufficiently  great  to  enable  the  eye  to  command  the 
whole  surface  with  perfect  convenience,  the  represen- 
tation, though  strictly  according  to  the  mathematical 
rules  of  perspective,  will  be  deceitful.  A rule  to  deter- 
mine this  distance  is  therefore  required,  and  it  will  be 
found,  that  the  eye  should  never  approach  nearer  the 
perspective  plane  than  to  a distance  equal  to  the  diagonal 
of  the  picture. 

For  example,  let  CDBA  (PI.  21,  Fig.  1),  represent 
the  picture,  then  the  distance  of  the  point  of  sight  from 
its  surface  should  not  be  less  than  AD.  It  .may  be 
greater,  and  this  is  left  to  the  taste  of  the  artist. 

105.  Let  it  be  required  to  put  a pavement  in  perspec- 
tive, composed  of  alternate  squares  of  black  and  white 


158 


TREATISE  ON 


marble — the  squares  of  equal  size.  Let  ABDC  (PL  21, 
Fig.  2),  represent  the  picture;  EF  its  horizon,  and  G 
the  centre  of  the  picture ; and  let  the  distance  of  the 
eye  from  the  perspective  plane  be  equal  only  to  GH. 
Let  us  now  take  the  same  example  (PI.  21,  Fig.  3),  and 
let  the  distance  from  the  point  of  sight  to  the  perspec- 
tive plane  be  equal  to  the  diagonal  of  the  picture. 

It  is  evident  that,  in  the  first  example,  the  pieces  of 
which  the  pavement  is  composed  neither  appear  to  be 
squares,  nor  of  equal  size,  while  in  the  second  figure 
they  do. 

106.  The  vanishing  point  of  perpendiculars,  or  the 
point  directly  opposite  the  eye,  has  been  called  the 
centre  of  the  picture.  This  term  has  been  used  merely 
to  designate  the  point,  and  does  not  imply  that  the  point 
must  be  situated  in  the  centre  of  the  canvass  in  which 
the  picture  is  drawn.  Indeed,  it  is  not  absolutely  neces- 
sary that  it  should  be  situated  within  the  limits  of  the 
canvass.  It  always,  however,  determines  the  horizon 
of  the  picture,  which,  by  a law  of  nature,  rises  and  falls 
with  the  eye. 

In  the  example  (PI.  21,  Fig.  4),  in  which  AB  is  the 
horizon,  the  centre  C,  of  the  picture,  lies  without’  the 
limits  of  the  canvass. 

In  the  example  (PI.  21,  Fig.  5),  it  lies  without,  and 
below  the  base  line  of  the  picture. 

107.  The  situation  of  the  horizontal  line  is  altogether 
left  to  the  taste  of  the  artist,  and  must  depend  on  the 
nature  of  the  subject  to  be  represented.  If  the  appear- 
ance of  magnitude  in  a building,  or  of  majesty  in  a 
human  or  ideal  figure  is  required,  the  choice  of  a 
very  low  horizon  is  the  best.  It  has  been  remarked 
that  the  “ Apollo  Belvidere  was  not  made  to  muse  in 
a valley.”  In  a picture,  he  can  only  be  placed  on  an 


LINEAR  PERSPECTIVE. 


159 


eminence,  by  allowing  the  horizon  to  be  at  his  feet  or 
below  them. 

On  the  contrary,  where  a rich  back-ground  to  figures 
is  required,  it  may  often  be  best  obtained  by  placing 
the  horizon  very  high  in  the  picture.  In  landscape 
painting,  the  most  usual  practice  is  to  place  the  horizon 
rather  below  than  above  the  middle  line  of  the  canvass. 

108.  Artists  are  accustomed  to  divide  linear  perspec- 
tive into  two  kinds,  parallel  and  oblique,  and  to  consider 
the  choice  of  either  as  entirely  a matter  of  taste.  The 
former  is  the  easiest,  and  therefore  from  indolence 
would  be  most  frequently  chosen.  Perhaps,  however, 
that  in  general  it  looks  the  best. 


THE  END. 


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